This is not so, or it is equally so, depending on how exactly you interpret things.

To get expectation values of either x or p, we need to multiply by x (or p), and integrate over the entire configuration space. In that sense, they are both non-local.

In order to apply the Schroedinger equation:

-i h d/dt (psi(x)) = -h^2 (d/dx)^2(psi(x))/2m + V(x)(psi(x))

we can act locally in the position basis: we only need to examine the area around x to update psi(x) in the next timestep.

Or we can look at it in the momentum basis:

-i h d/dt (psi(p)) = p^2(psi(p))/2m + V(i h d/dp)(psi(p))

This is exactly as local in the momentum basis as the position basis. We only need to look at the area around p to update psi(p) for the next timestep.

They really are on equal footing.

EDIT: There is one slight complication -- an infinite number of derivatives truly can become non-local: one nice example is exp(- i a p/h) psi(x) = exp(- a d/dx) psi(x) = psi(x-a). This is a reflection of momentum being the generator of displacement.