How would you prove that $$\oint_\Gamma \nabla\theta\cdot\vec{dr}=\pm2\pi $$
We know that $\theta\in(-\pi,\pi)$, suppose that $\theta$ is continuous in the region bounded by and along $\Gamma$ apart from a cut. On one side of the cut $\theta=-\pi$ and the other $\theta=\pi$, so $\theta$ makes a jump of magnitude $2\pi$ over this cut. Effectively I am describing the atan2 function, however i'm not assuming uniform change in $\theta$.
Is this as simple as saying, let us treat this contour as a disconnected line integral $\Gamma:[0,1]\rightarrow\mathbb{R}$ such that $\theta(\Gamma(0))=\mp \pi$ and $\theta(\Gamma(1))=\pm \pi$, therefore, $$ \int_\Gamma \nabla\theta\cdot\vec{dr}=\theta(\Gamma(1))-\theta(\Gamma(0))=\pm2\pi $$
I'm asking as it's been a while since i've studied any formal complex analysis and i'm hopeful there's a more formal way to think about this.
All you need is the Gradient theorem here. Since your $\Gamma$ is a line, with the ends almost meeting, it's like thinking of $(0,1)$ wrapped around in a circle missing the connecting piece. Once you see that, we have $$ \int_ \Gamma \nabla f \cdot dr = f(b) -f(a) $$ where $a$ and $b$ are the end points of $\Gamma$. The only thing you haven't specified is the orientation, this is where the $\pm 2 \pi$ comes in. Since it's either $$ f(b)-f(a) \quad \text{or} \quad f(a) -f(b) $$ depending on if you start from $a$ and go to $b$ or vice versa.
Ref: http://en.wikipedia.org/wiki/Gradient_theorem