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.