 Solution

No. The missing point is not assumed to complete $S$ as a smooth surface, so the hypotheses of the global Gauss-Bonnet theorem do not apply to $S\cup \{p\}$.

For a concrete counterexample, rotate a smooth simple arc $(r(s),z(s))$, parametrized by arc length for $0\le s<L$, about the $z$-axis. Arrange that it starts smoothly on the axis,stays in $r>0$ for $0<s<L$, and converges to another point of the axis as $s\uparrow L$, withThe endpoint is omitted. The resulting surface of revolution is diffeomorphic to ${\mathbb{S}}^2$ minus one point, and adjoining its limiting endpoint makes the set compact.

For an arc-length surface of revolution,Therefore the total Gaussian curvature of a punctured surface with a singular compactification point iswhich can take values different from $4\pi$. Compactness of the one-point closure alone therefore does not determine the integral.