What goes wrong with Stokes theorem if a surface is not orientable? For the Möbius strip parametrized by $\{\sigma(\theta,r)=((1+r\sin(\theta/2))\cos\theta,(1+r\sin(\theta/2))\sin\theta,r\cos(\theta/2))\ \mid \\ (\theta,r)\in A=(0,2\pi)\times(-1/2,1/2) \}$ 
we get the normal vector  $\nu$ by doing the cross product $$\nu(\theta,r)=\frac{\partial \sigma}{\partial \theta}\times\frac{\partial \sigma}{\partial r}=\begin{pmatrix}(1+r\sin{\theta\over2})\cos{\theta\over2}\cos\theta+{r\over2}\sin\theta \\(1+r\sin{\theta\over2})cos{\theta\over2}\sin\theta-{r\over2}\cos\theta \\-(1+r\sin{theta\over2})\sin{\theta\over2} \end{pmatrix}$$ which is continuous on $A$ but not on $\overline  A$ because $\nu(0,0)=(1,0,0)\ne(-1,0,0)=\nu(2\pi,0)$
For what kind of $F(x,y,z)=(F^1,F^2,F^3)$ with $F^i\in C^1(\Bbb R^3)$ can there be a problem? I lack some intuition for the necessity of the  orientability for Stokes' theorem:
$$\iint\limits_\Sigma \nabla \times F\cdot ds=\int\limits_{\partial\Sigma}F\cdot dl$$
 A: Edit: Just noticed that your strip does not include the points $\theta = 0, 2\pi$ which means that it can be oriented.
However, on the full domain you cannot define the vector surface integral to begin with and so the problem is not with Stoke's Theorem per say.Remember, that by definition a vector surface integral is:
$$
\iint_{\mathcal{S}} \overrightarrow{F} \cdot \, \mathrm{d}\overrightarrow{S} = \iint_\mathcal{S} \left(\overrightarrow{F} \cdot \hat{n}\right) \mathrm{d}S
$$
In other words you need a clear definition of $\hat{n}$ throughout $\mathcal{S}$. The Möbius strip has only one side and therefore cannot be oriented and that's why you cannot even define one side of equation for the Stoke's Theorem.
To get an intuition for why the Möbius strip cannot be oriented I suggest making one yourself. It's easy --just take a strip of paper and glue the two ends together by introducing a 180 degree twist somewhere in between. Now take a normal vector at a point of your choice anywhere on the surface. Translate this vector along the strip and arrive back at this point. Your normal vector will now be pointing in the reverse direction from when you started at that point.
This shows that you cannot define $\hat{n}$ consistently for this surface.
