Science nerd alert:
You have to be careful here to distinguish force and power, as well as different types of frictional (drag) forces.
Let's start with different frictional forces. We have:
1) Newtonian friction, where the frictional force is proportional to the speed (and a friction coefficient). F=eta*v, where F=force, v=velocity and eta=coefficient of friction. The type of friction which typically obeys this law is friction of mechanical parts of the bike or of tire on road surface, etc.
2) Air resistance, where the frictional force is proportional to the velocity squared (creating air turbulence in the wake), the effective surface area A, and some coefficient alpha. So we have F=alpha*A*v^2
For a flat, fast time trial, I think air drag is quite a bit larger than Newtonian friction, so one might neglect the latter and only consider the former.
Now, when talking about power to overcome the frictional force, one has to consider that P=F*v, i.e. that power is equal the product of the frictional force times the velocity. So, for air drag, power becomes proportional to velocity
cubed. (See for instance
Wikipedia)
Meaning that for otherwise equal conditions, power is proportional to effective area*velocity^3. (If your starting point is power/weight numbers, you'd have to multiply those with the weights of the riders to get numbers for power first.)
Let's assume rider A has 9% more power (not power/weight, just plain power) than rider B, but also 3% more effective area (he's bigger), then the increment in top velocity would be only one third of the difference, i.e. percent increase in velocity would be (9%-3%)/3=2% (This simple formula is only valid for small percent changes of area and power).
I think you made a mistake when you calculated the velocity ratios from the power ratios, and also, you forgot to take into account the difference in sizes (effective surface area) for the two riders. Based on that, I'd think relative to your prediction, Contador will do better (and Armstrong worse).
end science nerd alert