How are you measuring this 2V? From context it sounds like you may be using a multimeter rather than an oscilloscope. To really see what's going on in a circuit like this, the oscilloscope is the instrument of choice. You would be able to see from the symmetry of the charge and discharge curves that the the rates are indeed the same.
But it sounds like you are using a meter, and with a device that just gives you a number, understanding what's going on requires some interpretation.
It seems reasonable to interpret the input signal you described as a 5V peak-to-peak square wave riding on a 2.5V DC offset. So if you use a DC measuring device, you might expect to measure this 2.5V DC level across the capacitor.
If your measuring device happens to be a DVM, you can reasonably ignore the effects of the meter on the circuit. Even cheap digital meters have megohms of impedance and will not load down the k-ohm scale circuit under test. However, these kinds of meters vary widely in their ability to make sense of time-varying inputs. Some are good just for checking batteries. Some will give you a fair DC reading in the presence of sinusoidal AC, but not with more complex AC. Some will give you true RMS no matter what the shape of the waveform.
And if you are measuring using an old mechanical-movement type meter, you have to bear in mind that, as volt-meters, these meters are equivalent to a few k-ohms, maybe 10's of k-ohms at best. Hooking this kind of a meter into the circuit you describe will very certainly load the circuit and change its behavior significantly. You will get readings to be sure, but you have to interpret these knowing how the circuit is affected. In the case of the R-C setup you describe, this kind of meter would read lower than a DVM, on the basis that its resistance would help discharge the cap while contributing nothing to charging it.