In Graham's law, is the rate of effusion only dependent on molar mass? According to Graham's Law, the effusion rate of a gas is inversely proportional to the square root of the molar mass of the gas. 

However, consider this situation: We have s balloon with 1 mole of neon gas and another with 2 moles of neon gas. The gases are in thermal equilibrium and occupy the same volume. In this scenario, would the rate of effusion be the same?

The molar mass of the neon in both balloons is the same, so according to Graham's Law the rates would be the same. But intuitively, thinking about 2 moles versus 1, the pressure is doubled if volume and temperature are kept constant. And if the pressure is doubled, more collisions per second occur, so I would think that despite Graham's Law, the rate of effusion is also affected by the number of moles.
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.