Let $k \le n$ be the rank of $B$, and for $1 \le i \le k$, let $r_i(A)$ be the euclidean norm of the $i$th row of $A$ and let $c_i(A)$ be the euclidean norm of the $i$th column of $A$. Taking things from where you left off, one can write $$ \begin{split} \mbox{tr}(A^2 (A \circ B)) &\le \sum_i \sigma_i(A)^2 \sigma_i(A \circ B) \le \|A\|_{op}^2\cdot\sum_i\sigma_i(A \circ B)\\ &\le \|A\|_{op}^2\cdot \sum_i \min(r_i(A),r_i(A))\sigma_i(B)\\ &\le \|A\|_{op}^2\|B\|_\star\cdot\max_{1 \le i \le k}\min(r_i(A),c_i(A)), \end{split} $$ where $\|B\|_\star := \sum_i \sigma_i(B)$ is the nuclear norm of $B$ and the second line is by Theorem 1 of "INEQUALITIES FOR THE SINGULAR VALUES OF HADAMARD PRODUCTS".
Now, by standard RMT, we know that $\|A\|_{op} = \mathcal O(1)$ with probability $1-e^{-cn}$. Also, by standard arguments, each $r_i(A)^2$ and $c_i(A)^2$ have exponential concentration around mean, namely $1$. We deduce by a simple union bound that
$\mbox{tr}(A^2(A \circ B)) \lesssim \|B\|_\star$ w.h.p.