The Complexity of Advice in Computational Indistinguishability
Case 1. The distributions $U_n$ and $V_n$ are computable in the sense that we can eventually tell if your displayed inequality is violated, for fixed $n$ and $z$.

It seems the answer to the first question (following "The question is") is No in this case. Suppose there is a $D$ and a $p$ such that for infinitely many $n$, there is a $z$ violating the inequality. I claim we can find a computable function $f$ such that $f(1^n)$ is infinitely often such a $z$. What $f$ does is run all searches for $z$ for various values of $n$ in parallel. When one is found for some $n$, we define $f(1^n)=z$ and the searches for undefined values $m&lt;n$ are called off and $f(1^m)$ is defined arbitrarily. Then another parallel search is instituted for a larger value of $n$. 

Case 2, we do not make any assumption about $U$ and $V$. Then Yes. Both $U_n$ and $V_n$ will be deterministic in the following construction.

Let $D(1^n,w,z)$ simply compute whether $w=z$ and output 1 if this is the case. But $U_n$ and $V_n$ will output the $n$th element $s_n$, $t_n$, of sets $S$ and $T$, respectively. Then such a function $f$ would have the property that infinitely often $f(1^n)=s_n\ne t_n$ or vice versa. Now suppose that $n\mapsto s_n$ and $n\mapsto t_n$ both dominate all computable functions, and $\{n: s_n\ne t_n\}$ is infinite (such sets $S$ and $T$ clearly exist). Then a function $f$ exists with the required property, but cannot be computable.