Is every nonnegatively curved plane conformal to the complex plane? 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.
 A: Cheng-Yau proved that: A complete Riemannian manifold with non-negative Ricci curvature
and at most quadratic growth for volumes of balls as the radius goes to infinity is
parabolic.
EDIT (by Igor Belegradek). Various criteria for parabolicity are found in the survey of Grigoryan. In particular, on page 177 it is mentioned that a a complete Riemannian manifold with at most quadratic volume growth is parabolic. For complete open nonnegatively curved surfaces the volume growth is at most quadratic by Bishop-Gromov. On the other hand, there are parabolic complete manifolds with arbitrary fast volume growth (see page 180 of the same survey). Finally, the very first proof of parabolicity of complete nonnegatively curved plane seems to be due to Blanc-Fiala (1941); the reference is in Huber's paper mentioned in my comment to Anton's answer.
A: Assume your surface is conformally equivalent to a disc $D$ and $e^\phi$ be the conformal factor. From completeness, $\phi(x)\to\infty$ as $x\to \partial D$.
Gauss curvature can be expressed as $K=-\frac12{\cdot} e^{-\phi}{\cdot}\Delta\phi$.
Thus, $\Delta\phi\le 0$. 
The later contradicts maximum principle.
P.S. As Igor noticed, the argument has a gap: we only have that upper limit of $\phi(x)$ is $\infty$ as $x$ converge to any point on the boundary. He also give a ref with a complete proof. The argument would work if for any superharmonic function $\phi$ on $D$ there is a curve $\gamma$ from $0$ to the boundary such that 
$$\int\limits_\gamma e^\phi<\infty.$$
The later is proved by Fedja Nazarov here.
