The following is a theorem of [Elkies](http://www.ams.org/mathscinet-getitem?mr=903384):


Let $X$ be an elliptic curve over $\mathbb{Q}$. Then there are infinitely many primes $p$ such that the action of Frobenius on $H^1(\mathcal{O}, X)$ is zero.


Allen Knutson and I would like a similar theorem for higher dimensional Calabi-Yau varieties. Unfortunately, we've been told that this is probably open. (References for the fact that it is open are appreciated.) But we don't need the full strength of Elkies' result. It would be enough for us to know the following:


Let $X$ be an $n$-dimensional, smooth, complete Calabi-Yau variety over $\mathbb{Q}$, for $n>0$. Write $X/p$ for the fiber of $X$ over $p$. Let $T(p)$ be the action of Frobenius on $H^n(\mathcal{O}, X/p)$. Are there infinitely many $p$ for which $T(p) \neq 1$? Also, in the same generality, are there infinitely many primes for which $T(p) \neq 0$?



Elkies' supersingularity theorem in higher dimension
If I understand your question correctly, the "easy" version of the question you ask is unknown in dimension $\ge 3$, and is basically easy for surfaces of interest.


These questions are certainly unknown in dimension $\ge 3$, or else the Sato-Tate conjecture for weight $>3$ would have been known $12$ months ago, rather than $4$ months ago.


I take it you are only asking that the action of Frobenius is neither almost always trivial ($T(p) = 0$) or acts as the identity ($T(p) = 1$), which is much weaker than asking that there exist any supersingular primes.
Assume otherwise. Fix a prime $\ell$, and Let $V$ be the $\ell$-adic etale cohomology
$H^2(X)$ with the usual action of $\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$. Let $F\_p(T)$ denote the characteristic polynomial of Frobenius. It has coefficients in $\mathbf{Z}[x]$. Say one wants to show that there exist infinitely many primes $p$ such that $T(p) \ne 0$. You are imposing the condition that $F\_p(T) \equiv T^n \mod p$ for all but finitely many $p$, where $n = \mathrm{dim}(V)$. 


Choose a prime $L > 2n$. There will a positive density of primes such that $F\_p(T) \equiv (T-1)^n \mod L$.
If $p \equiv 1$ modulo $L$, These conditions together imply that the trace of $F\_p(T)$, which is *a* *priori* an integer, is $np \mod pL$. By the Weil conjectures, the roots of $F\_p(T)$ have absolute value $p$, and thus, since $L > 2n$, they must all equal $p$. By Cebatarev density, it follows that a finite index subgroup of the Galois group acts (on the semisimplification of $V$) via the cylotomic character on $V$. By looking at Hodge-Tate weights this implies that $h^{2,0} = h^{0,2} = 0$, and so (for example) one has a contradiction for a K3-surfaces.


If one wants to rule out that $T(p) = 1$ for (almost all) $p$, one is imposing the condition that $F\_p(T) = (T-1)^n \mod p$. In the same way one obtains an open subgroup of the Galois group such that the trace of Frobenius is always $n$, and then one deduces that $h^{0,2} > 0$ and $h^{2,0} = 0$, which can never happen.


EDIT: Forgot to mention that it was Ogus who proved in ~70's that Abelian surfaces had infinitely primes of ordinary reduction, presumably by a very similar argument. Well, except for the comparison theorem of Faltings I used above...


