$\newcommand{\F}{\mathbb{F}} \newcommand{\End}{\mathrm{End}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}}$ I would like to know if the following is true:

Proposition A : Let $A_1, A_2$ be two abelian varieties over a finite field $k$. If $\End_{\overline k}(A_1) \otimes_{\Z} \Q$ and $\End_{\overline k}(A_2) \otimes_{\Z} \Q$ are isomorphic (as $\Q$-algebras), then $A_1, A_2$ are isogenous over $\overline k$.

The converse holds: let $\phi : A_1 \to A_2$ be an isogeny of degree $m$. There is an isogeny $\psi : A_2 \to A_1$ such that $psi \circ \phi = [m]$ (see Poonen "Lectures on rational points", Proposition 4.1.19 ; this is not the dual isogeny!). Define a map $\End_{\overline k}(A_1) \otimes_{\Z} \Q \to \End_{\overline k}(A_2) \otimes_{\Z} \Q$ via $f \longmapsto \dfrac{1}{m} \phi \circ f \circ \psi $. It is an algebra isomorphism.

The result holds over $\mathbb C$, see here or Prop. 1.2.17. It holds for elliptic curves over a finite field: the supersingular case can be proved "by hand", using Tate isogeny theorem, while the ordinary case follows from Deuring's work on CM (lifting, etc.). Note that the curves do not need to be isogenous over the base field $k$. I skimmed through the paper Endomorphisms of Abelian Varieties over Finite Fields of Tate, but I did not find such a statement.

Maybe one could use Serre--Tate theory instead of Deuring, but it seems to be only available for ordinary abelian varieties. I am not aware of the details here anyway. If the claim does not hold, is there a suitable hypothesis to make it true?