(I assume by a plane field you mean a distribution.)

No, take $\mathbb{T}^3$ parametrized as $(x,y,z) \in [0,2\pi)^3$. The field $v = \partial_x + \sin(z) \partial_y$ is geodesic. But for any $\phi(x,y,z)$ the deformation tensor $\mathcal{L}_{\phi v} g$ can be computed to be
$$ \begin{pmatrix}
2 \phi_x & \phi_y + \phi_x \sin(z) & \phi_z \newline
\phi_y + \phi_x \sin(z) & 2\phi_y \sin(z) & \phi_z \sin(z) + \phi \cos(z) \newline
\phi_z & \phi_z \sin(z) + \phi \cos(z) & 0 \end{pmatrix} $$
which can only vanish if $\phi_x = \phi_y = \phi_z = 0$ (from the (1,1), (2,2) and (1,3) components) and $\phi \equiv 0$ (from the (2,3) component).

For distributions of higher dimension, you can observe that for any complete Riemannian manifold $TM$ as a top-dimensional distribution is geodesic. So it suffices to find a Riemannian manifold of the appropriate dimension that does *not* admit any Killing vector field to get a counterexample. Alternatively you can also modify the above one dimensional construction in the obvious way.