To start with I want to clear up an apparent misconception. The virial theorem is usually applied to systems in dynamical equilibrium, i.e. averaged quantities like the total kinetic energy don't evolve. This does not mean that there is no net force on any part of the system: systems of orbiting particles can be in dynamical equilibrium, and they obviously need forces to accelerate them along their orbits.
The thing you're calling the virial theorem (not incorrectly, most introductory texts do the same) is actually a special case of the full virial theorem:
$$\frac{1}{2}\frac{{\rm d}^2I}{{\rm d}t^2} = 2T + U + \Sigma$$
$I$ is the moment of inertia, $T$ is the kinetic energy of the system, $\Sigma$ is the work done by any external pressure and $U$ is the potential energy of the system (if external masses can be ignored in the calculation of the potential). The equation you stated I would normally call 'the condition for virial equilibrium'.
If $\Sigma$ is negligible (which is assumed in the form you give), then a system which has $2T < -U$ will have a dynamical evolution that drives an increase in $I$, or in other words the system contracts, and vice-versa for the opposite case. The system can also respond in other ways, for instance it might shed some $T$ via radiation.
If $2T\sim-U$, then $\frac{{\rm d}^2 I}{{\rm d}t^2}=0$, so the expansion/contraction is neither accelerating nor decelerating. This is a necessary condition for equilirbium, but it is not the same as $\frac{{\rm d}I}{{\rm d}t}=0$: the system might still be expanding/contracting and be about to 'overshoot' its equilibrium point. For such an overshoot to happen requires $\frac{{\rm d}T}{{\rm d}t}$ and $\frac{{\rm d}U}{{\rm d}t}\neq 0$. A zero time average implies that these derivatives are zero and that the system is actually 'stuck' on its equilibrium (or oscillating around it if the time average is over a long enough window, which could sloppily be called an equilibrium system; this might be fine, depending on context).
