In natural units, where $hbar = c = 1$, what is $G$?

If $hbar = c = 1$ then length is time, and mass is inverse length or inverse time.

This is correct. In this way, since $[G]=[M^{-1}L^3T^{-2}]$ in SI units, its dimensionality reduces to $$ [G] =[M^{-1}L^3T^{-2}] =[M^{-1}M^{-3}M^{2}] =[M^{-2}] . $$ This tells you that there is a product of powers of $hbar$ and $c$ such that $G$ becomes an inverse-squared mass when you divide by it, and this product turns out to be $hbar c$. This then gives us the correct expression: $$ G = hbar c times 6.7times 10^{-57} (mathrm{eV}/c^2)^{-2}, $$ as a valid identity in SI units, which then reduces to $$ G = 10^{-57} mathrm{eV}^{-2} $$ if you set $hbar = c = 1$.