Let $E$ be a globally generated vector bundle of rank $r$ on a normal irreducible projective variety $X$. Suppose that $E$ induces a finite map
$$X \to \mathbb{G}r (H^0(E), r)$$
to the Grassmannian of rank r quotients.
Can we say that a symmetric power of $S^mE$ for $m>>0$ induces an immersion
$$X \hookrightarrow \mathbb{G}r (H^0(S^mE), R)$$
where $R = \binom{m+r-1}{m} = \mathrm{rk} S^m E$?


In case $r=1$ it is straightforward, as such a line bundle would be ample,
here we ae not supposing $E$ to be ample, the hypothesis imply only that $E$ is nef and $\det E$ is ample.


A positive answer would be equivalent to say, under the same hypothesis for $E$, that for $m >>0$ the image $V$ of the determinant map
$$ \Lambda^R H^0(S^m E) \twoheadrightarrow V \subseteq H^0 (\det S^m E)$$ separate points and tangent vectors of $X$.



Global sections of determinant bundle of symmetric powers
The suggestion in my comment turns out to be impossible, and this leads to counterexamples to the OP's question, already for $\text{dim}(X)=2$. Let $U$ be a $2$-dimensional vector space over a field $k$. Let $\mathbb{P}(U^\vee)$ denote $\text{Proj} \text{Sym}^\bullet\_k U$, i.e., $\mathbb{P}^1\_k$. To be very precise, there is a short exact sequence on $\mathbb{P}(U^\vee)$,
$$ 0 \to (\bigwedge^2\_k(U^\vee))^\vee\otimes\_k \mathcal{O}(-1) \xrightarrow{\alpha^\dagger} U\otimes\_k \mathcal{O}\_{\mathbb{P}(U^\vee)} \xrightarrow{\beta} \mathcal{O}(+1) \to 0,$$
and the surjection $\beta$ is universal in the usual sense.


Let $V$ be $\text{Syt}\_k^2(U)$, the $k$-subspace of $U\otimes\_k U$ of symmetric tensors (the same as the symmetric quotient $\text{Sym}\_k^2(U)$ except in characteristic $2$). Then $V$ is a $3$-dimensional $k$-vector space. Let $\mathbb{P}(V^\vee)$ be $\text{Proj}\text{Sym}\_k^\bullet V$, i.e., $\mathbb{P}^2\_k$. To be very precise, on $\mathbb{P}(V^\vee)$ there is a short exact sequence, $$ 0 \to F^\vee \xrightarrow{\gamma^\dagger} V\otimes\_k \mathcal{O}\_{\mathbb{P}(V^\vee)} \xrightarrow{\delta} \mathcal{O}(1),$$
and the surjection $\delta$ is universal in the usual sense.


Consider $X=\mathbb{P}(U^\vee) \times\_k \mathbb{P}(U^\vee)$, i.e., $\mathbb{P}^1\_k\times\_k \mathbb{P}^1\_k$, with its two projection $\pi\_1,\pi\_2: X \to \mathbb{P}(U^\vee)$. There is a surjection,
$$ \pi\_1^\*\beta\otimes \pi\_2^\*\beta: (U\otimes\_k U)\otimes\_k \mathcal{O}\_X \twoheadrightarrow \pi\_1^\*\mathcal{O}(+1)\otimes\_{\mathcal{O}\_X}\pi\_2^\*\mathcal{O}(+1).$$ Computing locally, the restriction of $\pi\_1^\*\beta\otimes\pi\_2^\*\beta$ to the subsheaf $$\text{Syt}^2\_k(U)\otimes\_k \mathcal{O}\_X\subset (U\otimes\_k U)\otimes\_k \mathcal{O}\_X$$ is still surjective, i.e., there is an induced surjection,
$$
\text{Syt}^2\beta:\text{Syt}^2\_k(U)\otimes\_k \mathcal{O}\_X \twoheadrightarrow \pi\_1^\*\mathcal{O}(+1)\otimes\_{\mathcal{O}\_X}\pi\_2^\*\mathcal{O}(+1).$$ 
Thus, there exists a unique morphism $f:X\to \mathbb{P}(V^\vee)$ such that $f^\*\delta$ equals $\text{Syt}^2\beta$. Of course this is nothing other than the usual degree $2$ morphism $\mathbb{P}\_k^1\times\_k \mathbb{P}\_k^1 \to \mathbb{P}\_k^2$, but with due attention paid in characteristic $2$.


Now consider the dual of the short exact sequence on $\mathbb{P}(V^\vee)$, i.e.,
$$ 0 \to \mathcal{O}(-1) \xrightarrow{\delta^\dagger} V^\vee\otimes\_k \mathcal{O}\_{\mathbb{P}(V^\vee)} \xrightarrow{\gamma} F \to 0. $$
The surjection $\gamma$ is also universal, thus identifying $\mathbb{P}(V^\vee)$ with the Grassmannian, $\mathbb{G}r(2,V)$. Define $E$ to be $f^\*F$ on $X$. Consider the restriction of $F$ to points and lines in $\mathbb{P}(V^\vee)$.


For a point of $\mathbb{P}(V^\vee)$ corresponding to a rank $1$ quotient, $V\twoheadrightarrow Q\_1$, the fiber $F|\_{[Q\_1]}$ is simply the dual of the $2$-dimensional kernel of the quotient. For a pointed line in $\mathbb{P}(V^\vee)$ corresponding to a flag of quotients, $V\twoheadrightarrow Q\_2\twoheadrightarrow Q\_1$, the restriction of $F$ to the line is isomorphic to $\mathcal{O}\oplus \mathcal{O}(1)$, but where the fiber of the positive summand at the marked point $[Q\_1]$ corresponds to the annihilator of the kernel of the surjection to $Q\_2$. In particular, this depends on the choice of the line containing the specified point $[Q\_1]$. As we hold $Q\_1$ fixed and vary the line, this image in $F|\_{[Q\_1]}$ of the positive summand varies over all $1$-dimensional subspaces of the $2$-dimensional fiber.


**Claim.** For every integer $m\geq 0$, the locally free sheaf
$F\_m := \text{Sym}^m\_{\mathcal{O}}(F)\otimes\_{\mathcal{O}}\mathcal{O}(-1)$ on $\mathbb{P}(V^\vee)$ has only the zero section. 


**Proof of Claim.**
Since the restriction of $F$ to a pointed line is isomorphic to $\mathcal{O}\oplus \mathcal{O}(1)$, also 
the restriction of $F\_m$ to every pointed line $(L,[Q\_1])$ is isomorphic to a direct sum, $$\mathcal{O}(-1)\oplus \mathcal{O}\oplus \mathcal{O}(+1)\dots \oplus \mathcal{O}(m).$$ In particular, the subsheaf spanned by global sections is a locally free subsheaf whose quotient is an invertible sheaf. For a specified point $[Q\_1]$ of the line, the fiber at $[Q\_1]$ of the "spanned subsheaf" is a codimension $1$ linear subspace $S\_L$ of the fiber $F\_m|\_{[Q\_1]}$. By the same analysis as in the previous paragraph, for fixed $[Q\_1]$, for every nonzero element $v$ of $F\_m|\_{[Q\_1]}$, there is a choice of line $L$ containing $[Q\_1]$ such that $v$ is not contained in $S\_L$. Since the fiber $v$ of a global section of $F\_m$ at $[Q\_1]$ is contained in $S\_L$ for every choice of $L$, it follows that every global section of $F\_m$ vanishes at $[Q\_1]$. Since this holds for every point, and since $\mathbb{P}(V^\vee)$ is reduced, etc., it follows that $F^m$ has only the zero global section. 


**QED Claim.**


Now consider the locally free sheaf on $X$, $f^\*\text{Sym}^m\_{\mathcal{O}}(F) \cong \text{Sym}^m\_{\mathcal{O}}(E)$. The global sections on $X$ of $f^\*\text{Sym}^m\_{\mathcal{O}}(F)$ are the same as the global sections on $\mathbb{P}(V^\vee)$ of $f\_\*(f^\*\text{Sym}^m\_{\mathcal{O}}(F))$.  

By the projection formula, this is canonically isomorphic to $\text{Sym}^m\_{\mathcal{O}}(F)\otimes\_{\mathcal{O}}f\_\*\mathcal{O}\_X$. However, $f\_\*\mathcal{O}\_X$ is canonically isomorphic to $\mathcal{O}\_{\mathbb{P}(V^\vee)}\oplus \mathcal{O}(-1)$, where the summand $\mathcal{O}\_{\mathbb{P}(V^\vee)}$ is just the image of $f^\#$. Since, as proved in the claim, $\text{Sym}^m\_{\mathcal{O}}(F)\otimes\_{\mathcal{O}}\mathcal{O}(-1)$ has only the zero global section,
it follows that the pullback map on global sections,
$$ f^\*: H^0(\mathbb{P}(V^\vee),\text{Sym}^m\_{\mathcal{O}}(E)) \to H^0(\mathbb{P}(U^\vee)\times\_k \mathbb{P}(U^\vee), \text{Sym}^m\_{\mathcal{O}}(f^\*E)),$$
is an isomorphism. Therefore, for every induced morphism as in the OP's question, the morphism factors through $f$. So the induced morphism can never be a closed immersion.


