Here is a counterexample to Proposition A provided by 8-dimensional abelian varieties $A_1$ and $A_2$ over a finite field $F_{p^2}$ where $p$ is any prime that is congruent to $1$ modulo $17$ (e.g., $p=103$). The corresponding endomorphism algebra is the $17$th cyclotomic field $E=Q(\zeta_{17})$. The congruence condition means that $p$ splits completely in $E$. It is known that $E/Q$ is a cyclic extension, the class number of $E$ is $1$ (see ``Introduction to cyclotomic fields" by Larry Washington) and any proper subfield of $E$ is totally real (because $17$ is a Fermat prime).

Let $O_E$ be the ring of integers in $E$ and $\iota: E \to E$ be the complex conjugation, which is the only element of order 2 in the cyclic Galois group $G:=Gal(E/Q)$ of order 16=2^4. (In particular, every nontrivial subgroup of $G$ contains $\iota$.)

Let S be the 16-element set of maximal ideals $\mathfrak{P}$ of integers $O_E$ with residual characteristic $p$. The group $G$ acts freely transitively on the set $S$ of maximal ideals in $O_E$ and $\Pi_{\mathfrak{P}\in S}\mathfrak{P}=p \cdot O_E$. Let H be the set of ideals $\mathfrak{B}$ of $O_E$ such that $\mathfrak{B}\cdot \iota(\mathfrak{B})=p \cdot O_E$. The set $H$ has $2^8=16^2$ elements. I claim that the natural action of $16$-element $G$ on $H$ is free. Indeed, if it's not free then there is $\mathfrak{B}\in H$ such that $\iota(\mathfrak{B})=\mathfrak{B}$ and therefore
$$p \cdot O_E=\mathfrak{B}\cdot \iota(\mathfrak{B})=\mathfrak{B}^2,$$
which implies that $p$ is ramified in $E$, which is not the case. So, the action is free and therefore $H$ consists of $16$ orbits of $G$.

Now let's construct Weil's $p^2$-numbers $\pi_1$ and $\pi_2$, using $\mathfrak{B}_1, \mathfrak{B}_2 \in H$ that belong to different orbits of $G$. Let $z_1, z_2 \in O_E$ be generators of ideals $\mathfrak{B}_1$ and $\mathfrak{B}_2$ respectively. Then both
$v_1=z_1 \iota(z_1)$ and $v_2=z_2 \iota(z_2)$ are ``real" (i.e., $\iota$-invariant) generators of $p\cdot O_E$, i.e., there are exist units $u_1,u_2 \in O_E^*$ such that
$$v_1=p u_1, \ v_2=p u_2.$$
Clearly, both $u_1$ and $u_2$ are real totally positive. Now let us put
$$\pi_1=z_1^2/u_1 \in O_E, \ \pi_2=z_2^2/u_2\in O_E.$$
Clearly,
$$\pi_1\cdot \iota(\pi_1)=p^2=\pi_2\cdot \iota(\pi_2)$$
(recall that $\iota(u_1)=u_1$ and $\iota(u_2)=u_2$). Taking into account that
$$\pi_1 O_E=\mathfrak{B}_1^2, \pi_2 O_E=\mathfrak{B}_2^2,$$
we conclude that $\pi_1$ and $\pi_2$ are not Galois-conjugate (and the same is true for powers $\pi_1^m$ and $\pi_2^m$ for any positive integer $m$). Clearly,
$$\pi_1 \ne \iota(\pi_1), \ \pi_2 \ne \iota(\pi_2)$$
and therefore
$$Q(\pi_1)=E= Q(\pi_2).$$
Now if $A_1$ (resp. $A_2$) is a simple abelian variety over $F_{p^2}$ attached (by Honda-Tate) to $\pi_i$ then the center of the division algebra $End(A_i)\otimes Q$ is isomorphic to $E$. Since $\pi_1$ and $\pi_2$ (and even their powers) are not Galois-conjugate then $A_1$ and $A_2$ are not isogenous over $F_{p^2}$ (and even over its algebraic closure). On the other hand, both $A_1$ and $A_2$ are obviously ordinary. Since they are simple, their endomorphism algebras are commutative, i.e., coincide with their centers and therefore both $End(A_1)\otimes Q$ and $End(A_2)\otimes Q$ are isomorphic to $E$, Hence,
$$\dim(A_1)=[E:Q]/2=\dim(A_2),$$
i.e., both $A_1$ and $A_2$ are $8$-dimensional and
$End(A_1)\otimes Q$ is isomorphic to $End(A_2)\otimes Q$.

Let me stress that both $A_1$ and $A_2$ remain simple over an algebraic closure of $F_{p^2}$ and their endomorphism algebras remain isomorphic to $E$.