Let $K$ be a number field and $A/K$ an Abelian variety; let $l$ be a (rational) prime. Do there exist infinitely many primes $\mathfrak{p}$ of $K$ such that the Frobenius at $\mathfrak{p}$ acts semisimply on the *integral* Tate module $$T_l(A)\otimes \overline{\mathbb{Z}_\ell}?$$
By this I mean that there is a $\overline{\mathbb{Z}_\ell}$-basis of eigenvectors for the Frobenius action.

The answer is well-known to be yes if the integral Tate module is replaced with the rational Tate module (this goes back to Weil). If this is true, it should probably boil down to some Chebotarev argument, but I'm blanking at the moment.

Some remarks: As znt notes, the answer is "yes" if the image of the representation $G_K\to GL(A[\ell])$ contains a matrix with distinct eigenvalues, by Chebotarev.

Moreover, one can show that for any prime $\mathfrak{p}$ at which $A$ has good reduction, the Frobenius at $\mathfrak{p}$ acts semisimply on the $\ell$-adic Tate module for almost all $\ell$ (this follows from the theory of Frobenius tori). If the Mumford-Tate conjecture holds for $A$, I think it's enough to find any torus with good reduction at $\ell$ in the Mumford-Tate group of $A$, though I haven't written out the details. In any case, to find a counterexample it should be enough to find an Abelian variety over $K$ whose Mumford-Tate group contains no torus with good reduction at $\ell$.

One may also show the result holds if the image of the map $G_k\to GL(T_\ell(A))$ contains any element with distinct eigenvalues which is semi-simple over $\overline{\mathbb{Z}_\ell}$.