A reformulation of the problem, not yet an answer: Since the weight of $A$ in the statistical average can be arbitrarily small, we might as well replace $\mathbb{E}(A)$ by an arbitrary PSD matrix $B$ with $\lambda_2(B)>0$. The problem in the OP then amounts to the inequality 
$$|\lambda_2(A)-\lambda_2(B)|\leq \|A-B\|,$$
which remains to be proved.