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"][1].

Now, by gaussian concentration, both $r_i(A)$ and $r_i(A)$ have exponential concentration around $1$. We by a simple union bound that for any $\epsilon>0$, it holds w.p $1-ke^{-\Omega(\epsilon^2 n)}$ that
$$
\mbox{tr}(A^2(A \circ B)) \le \|A\|_{op}^2\|B\|_{\star}(1+\epsilon).
$$

Written more compactly,

>$\mbox{tr}(A^2(A \circ B)) \lesssim \|A\|_{op}^2\|B\|_\star$ w.h.p.


  [1]: https://epubs.siam.org/doi/abs/10.1137/S0895479896309645?journalCode=sjmael