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][1].
  


  [1]: https://mathoverflow.net/questions/58767/curve-integral-of-exponent-of-superharmonic-function/66501#66501