I think I have proved the following version of Stokes' theorem:
Teorem 1: Let $\beta: [0,4] \to \mathbb{R}^2$ the curve given by \begin{equation} \beta(t) = \begin{cases} (t,0) & \mbox{if } t\in[0,1], \\ (1,t-1) & \mbox{if } t\in[1,2], \\ (3-t,1) & \mbox{if } t\in[2,3], \\ (0,4-t) & \mbox{if } t\in[3,4]. \end{cases} \end{equation} This is the boundary of the square $D = [0,1] \times [0,1]$, oriented counterclockwise with speed 1. Suppose $A \subseteq \mathbb{R}^2$ is an open set containing $D$. If $\sigma: A \to \mathbb{R}^3$ is of class $C^2$, $\vec{F}: \mathbb{R}^3 \to \mathbb{R}^3$ is of class $C^1$ and $\alpha = \sigma \circ \beta$, then \begin{equation}\label{E:Stokes} \int_{\alpha} \vec{F} \cdot d\vec{r} = \iint_\sigma \mathrm{curl}\,\vec{F} \cdot d\vec{S}. \end{equation}
Apparently, I didn't need the usual hypothesis that $\sigma$ is a piecewise smooth parametrized surface. Is it in fact possible? Or I must have made some mistake?
Edit: My definition of the surface integral above is:
$$ \iint_\sigma \mathrm{curl}\,\vec{F} \cdot d\vec{S} = \iint_D \mathrm{curl}\,\vec{F}(\sigma(u,v)) \cdot \left( \frac{\partial \sigma}{\partial u}(u,v) \times \frac{\partial \sigma}{\partial v}(u,v) \right) \,dA. $$
And the smoothness condition is this:
http://mathworld.wolfram.com/SmoothSurface.html