Completeness implies that 
$$\int_{1/2}^1\sqrt{\rho(r,\theta)}dr=\infty$$
for all $\theta$.
So, for a complete metric,
$$\int_\Delta\sqrt{\rho}=\int_0^{2\pi}\int_0^1\sqrt{\rho(r,\theta)}rdrd\theta=\infty.$$
Thus $a\leq 1/2$.

For Poincare metric $\rho=1/(1-r^2)^2$, so
$\alpha=1/2$, and this is best possible.