This is a really good question ! Although Al Brown's answer is really good, I wanted to elaborate a bit further.
What is completely general is the Work-Energy Theorem which states that ∫𝐹𝑑𝑥 is equal to change in kinetic energy (this may be rotational or translational). In your question, you implicitly assumed that the friction doesn't exist. Without friction, we can write down the Newton's Equations of Motion for this particle (for sum of torques and forces):
F = ma (for linear motion) and FR = Ia/R (for rotational motion),
where I is the moment of inertia about the centre and R is the radius (Note that I have assumed the rotational acceleration to be equal to a/R which is not quite right as will be explained shortly). The second equation implies,
F = Ia/Rˆ2.
However, this is in contradiction with the first equation. I is equal to some factor times m*Rˆ2, so according to our second equation F = (some factor)*ma which may not always be right.
This happened because we have assumed that the ball was not slipping. This is not true because there is no friction force to balance this out ! This means that contrary to rolling without slipping, the point where the ball touches the ground is not stationary (in the case for rolling without slipping, the backwards angular motion and the forwards linear motion cancel at this point) and hence the ball will move more in a given amount of time due to the additional angular motion.
However, if there is friction and it is rolling without slipping, the ball will actually travel the same distance as a mass which can't roll. However, in this case, the work done by the force will be shared by the rotational kinetic energy and translational kinetic energy. A good example is the fact that in an inclined plane a full cylinder, sphere and a hollow cylinder doesn't arrive to the bottom at the same time.