Plotting the Phase Response
Your understanding is correct. There is no difference between a phase of $\pi$ and a phase of $-\pi$. You can always add or subtract integer multiples of $2\pi$ to the phase without changing anything because $e^{j2\pi n}=1$ for $n\in\mathbb{Z}$. Clearly you have $e^{j\pi}=e^{-j\pi}=-1$.

The probable reason why they used opposite signs in the phase plot for positive and negative frequencies is because then you can directly see from the figure that the phase (of a real-valued signal) is an  odd function, but that's a purely cosmetic matter.