No. Consider the very simple upper hemi-continuous correspondence $\phi$ from $[0,1]$ to $[0,1]$ such that
$$ \phi(x)= \begin{cases} \{0\} \text{ if }x<1/2\\ \{0,1\} \text{ if }x=1/2\\ \{1\} \text{ if }x>1/2. \end{cases} $$ Clearly, $[0,1]$ is an ANR. The correspondence $\phi$ looks almost like a discontinuous function, and that is why it cannot be well approximated by a continuous function. If there were an $\epsilon$-approximation $f_\epsilon$ with $\epsilon<1/4$, there must be some $x$ with $f_\epsilon(x)=1/2$ and there we have $d\big(f_\epsilon(x),\phi(x)\big)=1/2>\epsilon$.

No. Consider the very simple upper hemi-continuous correspondence $\phi$ from $[0,1]$ to $[0,1]$ such that
$$ \phi(x)= \begin{cases} \{0\} \text{ if }x<1/2\\ \{0,1\} \text{ if }x=1/2\\ \{1\} \text{ if }x>1/2. \end{cases} $$ Clearly, $[0,1]$ is an ANR. The correspondence $\phi$ looks almost like a discontinuous function, and that is why it cannot be well approximated by a continuous function. If there were an $\epsilon$-approximation $f_\epsilon$ with $\epsilon<1/4$, there must be some $x$ with $f_\epsilon(x)=1/2$ and there we have $d\big(f_\epsilon(x),\phi(x)\big)=1/2>\epsilon$.

It should be pointed out that almost the same argument works if we were to take $\phi(1/2)=[0,1]$. In that case, $\phi$ would be convex-valued but still no $\epsilon$-approximation as defined above possible. That is why one usually takes the difference between the graphs of the correspondence and the function as the relevant notion of approximation. The counterexample above is robust to this change.