 Solution

The outward area element is $d\mathbf{S}=a^2\sin \theta d\theta d\varphi {\mathbf{e}}_r$. ThereforeSplit the sphere into the northern cap $0\le \theta \le \alpha$ and the southern cap $\alpha \le \theta \le \pi$. Their induced boundary orientations are $+C$ and $-C$, respectively. The Stokes theorem applies to ${\mathbf{A}}^{(+)}$ on the north and ${\mathbf{A}}^{(-)}$ on the south. Adding their fluxes gives $2\pi (1-\cos \alpha )+2\pi (1+\cos \alpha )=4\pi$, agreeing with direct integration.

There is no contradiction with the flux of a globally smooth curl through a closed surface: neither potential is regular on the entire sphere. On their overlap they differ by $2{\mathbf{e}}_{\varphi }/(r\sin \theta )=2\nabla \varphi$, a local gauge transformation. The longitude is not a globally single-valued scalar, so that difference can have nonzero circulation around $C$.