Hardly. I am looking at the forces involved. The poster made a statement, that on a 700c bike, he was able to out-accelerate his partner who even though had 40 lbs more weight, still could not keep up with himself on a hill. I made an attempt to describe the forces involved. As far as total friction from the hubs, friction is a function of RPM as well as total friction. 700c wheels turn at a lower rpm than 26" wheels.

I used weight as a proxy for volume, and from that, extrapolating surface area, which is a good indicator of total aerodynamic drag, all else being equal. That is why I take the cube root of weight and square it. That will give us an idea of surface area, which is a good indicator of aerodynamic drag. Force of gravity is dependant on mass, which is density * volume. Total drag is a function of surface area, which has an exponential coeffecient of x^(2/3) that of volume. In any event, I am giving the lighter rider the benefit of the doubt in assuming that both riders are of equivalent physical condition, the only difference being size. A more out-of-shape rider is more dense than an in shape rider, meaning more mass compared to surface area.
No, it's just not that important. The force of aerodyamic drag is far, far more important than rotational inertia.

Except for the fact that weight *does* play a factor, or more accurately, the ratio of weight/aerodynamic drag plays a factor. A larger person has less surface area as a function of volume than a smaller person. All else being equal, total surface area is the greatest predictor of drag there is. It is not that a larger person gets a better assist from gravity, it is that they get less of an effect from drag. The greater the difference in size, the greater an effect aerodynamic drag has on the lighter rider relative to the heavier one. (If you don't think volume/surface area matters, explain why a 10cm hailstone has a terminal velocity of 136mph, yet a 1cm hailstone only has a term velocity of 46)