In Graham's law, is the rate of effusion only dependent on molar mass?
Molar mass is just one of the factors affecting effusion. Graham's law states that everything else remaining same, the rate of effusion of a gas is inversely proportional to $\sqrt M$. In other words, Graham's law describes the relation of the rate of effusion with only one of the factors involved in determining the overall rate.  

For the general rate of effusion of a gas at pressure $P$, and the hole area $A$ at a temperature $T$, using kinetic molecular theory, it turns out to be: (derivation on Physics.SE)

$$r=\frac{PA}{\sqrt{2\pi MRT}}$$

Now, in your question, the balloon with two moles will have a greater pressure (more moles in the same volume) and therefore it will have a greater rate of effusion for the same hole area and temperature and molecular mass.