I feel experts might be able to answer this question immediately.
Let $G$ be a connected $\mathbb Q$-simple and $\mathbb Q$-isotropic algebraic group. 
Let $S$ be a maximal $\mathbb Q$-split torus of $G$ and  

let $T\supset S$ be a maximal torus defined over $\mathbb Q$. Let $T\_a$ be the maximal anisotropic subtorus of $T$. Is $T$ a direct product of $S$ and $T\_a$?



structure of maximal tori in semisimple algebraic groups
The Weil-restriction construction suggested by user89334 is $\mathbf{Q}$-simple but not absolutely simple. To give absolutely simple examples, consider $G = {\rm{SL}}\_n(D)$ for a central division algebra $D$ over $\mathbf{Q}$ with dimension $d^2>1$. 


A maximal split torus in $G$ is given by the diagonal torus $S$ in the evident subgroup ${\rm{SL}}\_n \subset G$. The centralizer $Z\_G(S)$ is equal to $$\{(d\_1,\dots,d\_n) \in \underline{D}^{\times}\,|\,\prod {\rm{Nrd}}(d\_j) = 1\}$$
where $\underline{D}^{\times} := {\rm{GL}}\_1(D)$, so $\mathscr{D}(Z\_G(S))={\rm{SL}}\_1(D)^n$. This derived group has center containing $S[d]$, and every maximal torus $T \subset Z\_G(S)$ is uniquely an isogenous product $S \cdot T\_0$ for a (necessarily anisotropic) maximal torus $T\_0 \subset \mathscr{D}(Z\_G(S))$. Such a maximal $T\_0$ is the maximal anisotropic subtorus of $T$ and must contain the central $S[d]$ in the derived group due to its maximality, so $T$ is never a direct product of $S$ and $T\_0$ since $d > 1$. Passing to (central) isogenous quotients of $G$ never makes that central overlap entirely go away (since $S[d]$ is not cyclic), so the phenomenon persists for all members of the (central) isogeny class.


