To answer your questions in short:
- Why does friction stop exerting a force?
It doesn't!
Static friction still works. The wheel is not slipping; it is rolling. At the contact point the wheel and surface stay together because of static friction. This point of the wheel doesn't slide over the surface. So no kinetic friction, but certainly static friction. If not, then how would you start your car? Your tires need grip (static friction) on the surface, and must not slip.
The quote you give does though mention constant velocity. That is the same as saying no acceleration and of course also no angular acceleration. When that is the case, which it will be after some time, all torques must sum up to zero. So, if friction is the only force that causes a torque, then friction must be zero. Friction is only making the wheel start speeding up it's rotation - it makes the wheel start to turn, when your car speeds up. But when the rotation is constant, there is no more friction - just like when the block is sliding with constant speed on the ice, or when the space shuttle is drifting at constant speed, there is no friction that brakes it.
- Am I mistaken that friction exerts a force whenever an object and its surface might be distanced from each other?
Yes, you are mistaken. If I lift a bag of potatoes from the floor, there is no friction.
But I know what you mean: what if you move the object in parallel along the surface. Still, ideally you can have a no-friction icy surface, but apart from that I guess not.
There is more than one approach one can use to calculate the work done by a contact force on a rigid object. All approaches are valid so long as you're careful about how you actually use your calculated work.
One way is to use the infinitesimal displacement of the point of application $d\vec l_\text{app}$:
$$W = \int \vec F \cdot d\vec l_\text{app} \tag{1}$$
In the absence of other forces (and forms of potential energy), this will correctly give you the quantity $1/2 \int v^2 dm$, i.e., the change in total kinetic energy.
In your object-down-the-ramp example, the static frictional force would do zero work.
Another way to approach the same problem is to use the center of mass displacement $d\vec l_\text{CM}$:
$$W = \int \vec F \cdot d\vec l_\text{CM} \tag{2}$$
In the absence of other forces (and forms of potential energy), this will give you the change in center-of-mass kinetic energy $\Delta K_\text{CM} \equiv \Delta\left(\frac{1}{2}mv_\text{CM}^2\right)$.
Yet another way is an extension of method #2, in which one considers the rotational aspects of the motion:
$$W = W_\text{CM} + W_\text{rot} = \int \vec F \cdot d\vec l_\text{CM} + \int \vec\tau \cdot d\vec\theta \tag{3}$$
This latter approach lends itself well to separating out work that causes changes in the center-of-mass kinetic energy and changes in the rotational energy of a rigid body:
$$\Delta K_\text{tot} = \Delta K_\text{CM} + \Delta K_\text{rot}$$
More info on the different definitions of work: Article by Sherwood
Best Answer
Work is force times distance. If there is no slip, the force of friction acts over a distance of 0. There is no work.
Gravity does work. As the cylinder rolls down the hill, it accelerates. It gains kinetic energy in two forms: translation and rotation.
Gravity would do the same work on an identical cylinder that slide down the same slope without friction. The kinetic energy of the two would be the same at each position.
The rolling cylinder would travel more slowly than the sliding cylinder. But it would also spin.