Is it true that any complete metric of nonnegative (Gauss) curvature on $\mathbb R^2$ is conformally equivalent to the standard Euclidean metric on $\mathbb R^2$?

**Remarks:** locally any Riemannian metric on a surface is conformally equivalent
to the standard $\mathbb R^2$, due to existence of isothermal coordinates. Thus a Riemannian metric defines a conformal structure on the surface, and the uniformization theorem says that a simply-connected open surface is conformally equivalent to the complex plane or the unit disk. Thus the question is whether a complete nonnegatively curved plane is conformally equivalent to the complex plane.