Suppose $X \in \mathbb{R}^{n \times d}$ is a random matrix where $n > d$. Given a matrix $A \in \mathbb{R}^{n \times n}$ such that $AX$ is a zero matrix in expectation, i.e., $\mathbb{E}_{X}[AX] = 0$. Let $\sigma^2$ be the variance of the norm of $AX$, i.e., $\sigma^2:=\mathbb{V}[\lVert AX \rVert^2_F]$.
Now I would like to study the property of the random matrix $B=X(X^\top X)^{-1}X^\top AX$. Do we also have $\mathbb{E}[B] = 0$ and can we bound $\mathbb{V}[\lVert B \rVert^2_F]$ by $\sigma^2$?
$\newcommand\si\sigma\newcommand\bm[1]{\begin{bmatrix}#1\end{bmatrix}}$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 $Var\,\|B\|_F^2\le Var\,\|AX\|_F^2$.
No, you cannot conclude that $\mathbb{E}[B]=0$. Here is a counterexample for $n=2,d=1$: $A={{1\,1}\choose{0\,0}}$, $X={a\choose -b}$, with $\mathbb{E}[a]=\mathbb{E}[b]$. Then $AX={a-b\choose 0}$ satisfies $\mathbb{E}[AX]=0$, however the matrix $$B=X(X^\top X)^{-1}X^\top AX=\frac{a^2-ab}{a^2+b^2}{a\choose -b}$$ does not necessarily have a vanishing expectation value (for example, $\mathbb{E}[B]\neq 0$ if $\mathbb{E}[a^2b]\neq\mathbb{E}[ab^2]$).
