It appears to me that what happens depends on the smallest "Lie algebra" that contains the two vector fields $V = \nabla\phi_1$ and $W = \nabla\phi_2$. If the iterated Lie brackets of $V$ and $W$ generate a sub-bundle (i.e, the rank of the subspace at each point is a constant independent of the point), then that sub-bundle is integrable. Then $u$ most be constant along the integral submanifolds.

In any dimension it appears to me that what happens depends on the smallest "Lie algebra" that contains the two vector fields $V = \nabla\phi_1$ and $W = \nabla\phi_2$. If the iterated Lie brackets of $V$ and $W$ generate a sub-bundle (i.e, the rank of the subspace at each point is a constant independent of the point), then that sub-bundle is integrable. Then $u$ most be constant along the integral submanifolds.

ADDED: In dimension 3 there are only two nonsingular possibilities: Either $[V,W]$ always lies in the span of $V$ and $W$ or $[V,W]$ sis always transverse to the $2$-plane spanned by $V$ and $W$. In the first case, the span of $V$ and $W$ is integrable and $u$ can be any function that is constant along the integral surfaces. In the latter, $u$ must be constant on the whole manifold.

Other cases are more difficult to analyze.