The circuit with the non-ideal voltmeter will have less resistance so I presume the change of voltage would be slower, since the energy can flow more freely in the circuit without being damped quite so much.
Consider the extreme case where the the voltmeter is so bad that it essentially has no resistance and the current just flows right through it. This would be a short circuit, meaning the current would take that path without giving much energy at all to the resistor, which would mean overall the voltage would remain high for longer.
The circuit you've drawn wouldn't work. It would charge the capacitors, but when you close the switch to let them discharge, the positive charge on the plates will have no way of flowing round to the other side.
Have you done simple harmonic motion yet? Capacitors can actually be considered as SHM systems. When you let them discharge, the charge oscillates from positive, around the circuit, to negative, and back again, continuously until the energy is dissipated. So as you can see in your circuit diagram, there is no way for the charge to oscillate from the positive plates around to the negative plates - it's all just stored on one side and then stays there.
I've attached a circuit diagram that I think would work for what you wanted.