The answer is positive if we can assume that $U$ is Borel measurable (not necessarily open) and *its complement is negligible*. I will assume this in the sequel.

Fix Riemann metrics $g$ on $X$ and $h$ on $Y$. Denote by $k(p)$ the cardinality of $f^{-1}(y)$ and by $k_U(y)$ the cardinality of $f^{-1}(y)\cap U$. Denote by $J_f$ the Jacobian of $f$. More precisely for $x\in X$ we have

$$J_f(x)=\sqrt{\det D_xf\cdot(D_xf)^*}, $$

where $D_xf: T_x X\to T_{f(x)}Y$ is the differential of $f$ and $(D_x f)^*$ denotes its adjoint with respect to the inner products $g_x$ and $h_{f(x)}$.

The coarea formula shows that

$$ \int_Y k(y) dV_h(y)=\int_X J_f(x) dV_g(x)=\int_U J_f(x) dV_g(x). $$

The last equality follows from the fact that $X\setminus U$ has measure zero.

On the other hand the coarea formula (see Thm. 2.6 of this paper)shows that

$$\int_U J_f(x) dVg(x)=\int_Y|f^{-1}y)\cap U| dV_h(y)=\int_Y k_U(y) dV_h(y).$$

Thus

$$ \int_Y k(y) dV_h(y)= \int_Y k_U(y) dV_h(y) $$
or, equivalently

$$ \int_Y \Big(\; k(y)-k_U(y)\;\Big) dV_h(y)=0. $$

Since the measurable function $k(y)-k_U(y)$ is nonnegative we deduce from the above that $k(y)=k_U(y)$ for almost all $y$.