Suppose I have two non-CM abelian varieties $A$ and $B$ over $\mathbb{Q}$, both of dimension $>1$. Let $A_p$ and $B_p$ denote their reductions modulo $p$. If $A_p$ and $B_p$ are $\mathbb{F}_p$-isogenous for infinitely many primes $p$ (but not necessarily a positive density subset), must $A$ and $B$ be $\overline{\mathbb{Q}}$-isogenous?

EDITED: What if I instead require that both are $\mathbb{Q}$-simple (or perhaps geometrically simple if needed)? Or that $A$ and $B$ are modular, i.e, of $\text{GL}_2$-type?

I know the following special cases.

Suppose $A$ and $B$ are elliptic curves over $\mathbb{Q}$. This paper of François Charles proves that for any two elliptic curves $A$ and $B$ over $\mathbb{Q}$, their reductions modulo $p$ are $\overline{\mathbb{F}}_p$-isogenous for infinitely many $p$. I believe this is also expected to remain true if we replace $\overline{\mathbb{F}}_p$ with $\mathbb{F}_p$.

We can see this concretely for two CM elliptic curves as follows. Take $E/\mathbb{Q}$ and $E'/\mathbb{Q}$, where $E$ has CM by $K$ and $E'$ has CM by $K'$. Since we are in characteristic $0$, two CM elliptic curves are $\overline{\mathbb{Q}}$-isogenous if and only if their geometric endomorphism algebras are isomorphic, i.e. $K \cong K'$, so clearly $E$ and $E'$ are not $\overline{\mathbb{Q}}$-isogenous. However, for any prime $p$ (assume $p>3$) that is inert in both $K$ and $K'$ (and there are infinitely many such primes), $E_p$ and $E'_p$ are both supersingular modulo $p$, hence they are $\mathbb{F}_p$-isogenous.
Very strong multiplicity one for Hecke eigenforms shows that for non-CM elliptic curves $E$ and $F$ over $\mathbb{Q}$, if $E_p$ and $F_p$ are $\mathbb{F}_p$-isogenous for a density $1$ set of primes $p$, then $E$ and $F$ are $\mathbb{Q}$-isogenous. I believe user631's reply shows that if we replace density $1$ with positive density, then we can say that $E$ and $F$ are $\overline{\mathbb{Q}}$-isogenous.

Suppose $A$ and $B$ are CM abelian varieties of dimension $g>1$ over $\mathbb{Q}$, where $A$ has CM by $K$ and $B$ has CM by $K'$.

Since $K \not\cong K'$, $A$ and $B$ are not $\overline{\mathbb{Q}}$-isogenous. Consider primes $p$ that are inert in both $K$ and $K'$. There are infinitely many such primes, and for all of them, $A$ and $B$ both have supersingular reduction modulo $p$. Recall that there are only finitely many supersingular isogeny classes of $g$-dimensional abelian varieties over $\mathbb{F}_p$ for each $p$. I would assume that we have infinitely many primes $p$ in this subset for which $A_p$ and $B_p$ lie in the same supersingular $\mathbb{F}_p$-isogeny class, but I am unsure of a rigorous argument.

Suppose $A$ and $B$ are abelian varieties over $\mathbb{Q}$ such that $A_p \sim B_p$ for a density $1$ set of primes $p$. Then, I believe $A$ and $B$ are $\mathbb{Q}$-isogenous by Faltings' isogeny theorem and Chebotarev density.

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