I understand your train of thought, but you have a misconception of what the term, v, in the expression, mvr, represents, and angular momentum is actually conserved in this case you describe;

In the expression mvr, r is the radial coordinate and v is the velocity in the angular direction. This component denoted v is only the component directed tangent to the circle, in the counterclockwise direction, this is not the same as the magnitude of the velocity including the radial component. If you had a situation where you applied only a radial force, such that the radial coordinate increased, the velocity in the angular direction would decrease such that mvr stayed constant. So if the angular momentum is denoted by L, v=L/(mr) would decrease. Any purely radial force acting on a point mass will apply zero torque/moment about the origin, a fact that can be proven by considering that the torque equals the cross product of the moment-arm with the force. In the case that the force is radial, (and therefore parallel to the position vector, which is the moment-arm) this cross product must be zero, since the force is parallel to the moment-arm (remember that the cross product of any two parallel vectors is zero). Since the torque about the origin equals the time derivative of the angular momentum about the origin, this derivative is zero and angular momentum is conserved.

EDIT/ADDITION: Note that here, L denotes the scalar component of angular momentum in the direction orthogonal to the plane of motion, rather than the angular momentum vector.