Let $X=\beta\omega$, $a,b\in\beta\omega$ be two distinct free ultrafilters, $Y=X/\{a,b\}$ be the quotient space and $f:X\to Y$ be the quotient map. It is clear that $f$ is perfect and hence proper. 

Let $B=Y$, $A=B\setminus \{q(a)\}$, $h:B\to Y$ be the identity map, $g=q^{-1}|A:A\to X\setminus\{a,b\}\subset X$ be the homeomorphism of $A=Y\setminus\{q(a)\}$ onto $X\setminus\{a,b\}$. The set $A\subset B$ is ultrafilter-like, being open in $B$. It is easy to see that $f\circ g=h|A$. On the other hand, it can be shown that $g$ admits no continuous extension to a map $g':B\to X$ such that $f\circ g'=h$.