Why does the QCD vacuum have zero momentum?

It's by definition. A vacuum state is defined to be Poincaré invariant, since it should not depend on the frame (in special relativistic QFT; you get frame-dependent vacua in QFT in curved spacetime).

If it had non-zero momentum, it would not be invariant under rotations and boosts, for instance.

For the non-interacting vacuum, you can also easily see this: The vacuum is by definition the state that gives zero when any annihilation operator is applied to it - it is the "empty state". The mode expansion of the momentum operator of a non-interacting theory is $$ P^\mu = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}p^\mu a^\dagger(\vec p) a(\vec p)$$ (for a scalar field, neglecting a vacuum energy term for $P^0$), and so applying this to the non-interacting vacuum gives zero since the $a(\vec p)$ just give zero when acting on $\lvert 0 \rangle$.

@ACuriousMind 's answer is pretty straightforward. As an alternate proof consider the following:

• Note that $\hat P^\mu$ is time-independent$^1$, which means that it commutes with $\hat H$. So it commutes with $\exp[-i\hat HT]\ \forall T\in \mathbb C$.
• We know$^2$ that $|\Omega\rangle \propto \lim_{T\to\infty} \exp[-i\hat HT]|0\rangle$. Using this, its easy to see that $$ \hat P^\mu |\Omega\rangle\propto \lim_{T\to\infty} \hat P^\mu\mathrm e^{-i\hat HT}|0\rangle=\lim_{T\to\infty} \mathrm e^{-i\hat HT}\hat P^\mu|0\rangle=0 $$ where I used $[\hat H,\hat P^\mu]=0$ and $\hat P^\mu|0\rangle=0$.

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$^1$ here, $\hat P^\mu$ and $\hat H$ are the total (interacting) momentum-hamiltonian operators. They are time-independent because they are the charges of the Noether current associated to space-time translations.

$^2$ See Quantum Field Theory I, lecture notes by Timo Weigand, page 57.