No one answered this week's problem. You can read my solution below.
The answer is no. Suppose to the contrary that $\omega$ is nowhere vanishing. Then $\omega^2$, being a 4-form on a compact oriented 4-manifold has nonzero integral. The de Rham cohomology groups of the four-sphere are $\Bbb R$ in dimensions zero and four, and zero in every other dimension. Thus, $\omega$ is exact. If $\omega = d\eta$, then $\omega^2 = d(\eta \wedge d\eta)$. So $\omega^2$ is exact; its integral is zero by Stokes's theorem. This is absurd.