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 $f\circ g'=h$.

Let $C$ be a connected Tychonoff space and $a,b\in \beta C\setminus C$ be two distinct points. Let $X=\beta C$, $Y=X/\{a,b\}$ 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 $X\setminus\{a,b\}$. 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 $f\circ g'=h$.

The space $g(A)$ is connected sinse it contains a dense connected subspace $C$. Then the space $A$ is connected as well (being homeomorphic to $g(A)$). The connectedness of $A$ implies that $A$ is ultrafilter-like in $B$ (being connected the space $A$ admits no non-trivial partitions into two open sets).

So, to get a sensible answer, we should assume that the space $A$ is disconnected, or better (strongly) zero-dimensional. In this case there is a hope for the positove answer since ultrafilter-likenes of $A$ in $B$ should imply that the identity embedding $A\to\beta A$ extends to an embedding of $B$ into $\beta(A)$.