The short answer is an undoubted yes. The Higgs field (let's call it $\Phi$) is simply a scalar field, just like the ones you're used to seeing in standard QFT, which has a potential of the form
$$V(\Phi)=-\frac{1}{2}\mu^2\Phi^2+\frac{\lambda}{4}\Phi^4$$
Which gives your standard "Mexican Hat" potential. Of course, it is immediately seen that, even though the potential is symmetric under $\Phi\to-\Phi$, the two vacua (the points where $V$ is minimized) are $v=\pm\sqrt{\mu^2/\lambda}\,$, and the true vacuum can only be one of these. Let's assume it's the positive one. Then we define $\Phi=h+v$ and arrive at the potential
$$V(h)=\mu^2h^2-v\lambda h^3+\frac{\lambda}{4}h^4+C$$
Where $C$ is some irrelevant constant ($C=-\mu^4/4\lambda$, probably). Thus, we see that the Higgs particle (simply the original field $\Phi$ shifted by a constant) has mass $\sqrt{2}\,\mu$ and has cubic and quartic couplings (which are hopefully small). Because of this, it can be quantized in exactly the same way as any other scalar field, by considering the free theory and quantizing the fourier modes of that theory, which would look something like
$$h(x)\propto\int\frac{\mathrm{d}^4k}{(2\pi)^4}\left(a_{k}e^{ik\cdot x}+a^{\dagger}_{k}e^{-ik\cdot x}\right)\delta(k^2-2\mu^2)$$
And then treating the interactions as perturbations to a free Hamiltonian. So, yes, the Higgs boson does have a physical momentum, just like every other particle (that we know of).
While I'm not sure that I understand why you doubted that the Higgs had physical momentum in the first place, it's always good to ask questions.
EDIT: OP explained the confusion and I think I have an idea of what's going on here.
So, the Higgs field itself has a vacuum expectation value (vev), which we above called $v$. This means that the value of the Higgs field, so long as it is not perturbed, is exactly $v$ at every point in spacetime. While it is true that the Higgs field $h$ has physical momentum, the vacuum Higgs field $v$ does not. That is to say, that the vev does not carry any external momentum. It could, however, contribute to a cosmological constant since $V(v)$ does not necessarily vanish.
There are lots of little problems like this when you try to reconcile standard QFT with prediction of General Relativity. You can no longer shift your potential by an arbitrary constant (as that contributes to the stress-energy tensor), you cannot simply ignore vacuum energy, especially when it's infinite (this can be swept under the rug by including supersymmetry, though). It's just one of the many reasons why QFT and gravity don't like to play well.
