$\newcommand\si\sigma\newcommand\bm[1]{\begin{bmatrix}#1\end{bmatrix}}$No and no: In general, (i) $EB\ne0$ and (ii) we cannot bound $Var\,\|B\|_F^2$ by $\si^2$.

E.g., let $n=3$, $d=1$,
$$y_1:=\bm{1\\ -1\\ -1},\quad y_2:=\bm{-1\\ 1\\ -1},\quad
y_3:=\bm{-1\\ -1\\ 1},\quad y_4:=\bm{1\\ 1\\ 1},
$$
$$A:=\bm{1&-1&0\\0&1&0\\0&0&1}.$$
Let $Y$ be a random matrix such that $P(Y=y_j)=1/4$ for $j=1,2,3,4$. Let
$$X:=A^{-1}Y.$$

Then $EAX=EY=0$ and $\|AX\|_F^2=\|Y\|_F^2=3$ almost surely (a.s.), so that
$$\si^2=Var\,\|AX\|_F^2=0.$$

However,
$$EB=\bm{0\\ 0\\ -1/6}\ne0.$$
Also, the values of $\|B\|_F^2$ at $Y=y_1$ and at $Y=y_4$ are the non-equal numbers $2$ and $8/3$, respectively, so that $Var\,\|B\|_F^2$ is strictly greater than $0=\si^2$. (In fact, $Var\,\|B\|_F^2=1/9$.)

One might feel some affinity for the OP's conjectures. Indeed,

(i) we have $B=P_X AX$, where $P_X$ is the matrix of the orthoprojector onto the column space of the matrix $X$. So, if $P_X$ did not depend on $X$, the conjectured conclusion $EB=0$ would follow from $EAX=0$ and the linearity of any orthoprojector. Also,

(ii) since $P_X$ is the matrix of an orthoprojector, the equality $B=P_X AX$ does imply $\|B\|_F^2\le\|AX\|_F^2$. However, as shown above, this will not in general imply that $Var\,\|B\|_F^2\le Var\,\|AX\|_F^2$.