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 
$$\int\limits_\gamma e^\phi<\infty.$$
It is likely to be true, but I do not know enough to prove it.