The inequality $$ \mathbb E \|A G B\|_* \leq \min( \|A\|_* \|B\|_F,\|A\|_F \|B\|_*)$$ always holds. By symmetry, it is enough to prove the inequality $\mathbb E \|A G B\|_* \leq \|A\|_* \|B\|_F$. And by writing $A$ as a sum of rank one operators ("the unit ball of the trace class is the convex hull of the norm $1$ rank $1$ matrices"), we can assume that $A$ has rank one.
In that case, $A G B$ has rank one for every $G$. Using (1) that for a rank $1$ matrix, the trace norm and Frobenius norm coincide, and (2) that the $L^1$ norm is less than the $L^2$ norm, we get $$ \mathbb E \|A G B\|_* = \mathbb E \|A G B\|_F \leq (\mathbb E \|A G B\|_F^{2})^{\frac 1 2} = \|A\|_F \|B\|_F=\|A\|_* \|B\|_F.$$ The second equality is a straighforward computation, at least when $A$ and $B$ are diagonal.