when a resistor is added between them (diagram below), does that affect the conservation of charge and result in lower final voltages?
Think about it this way: your circuit is split into two halves by the capacitor dielectrics. On your schematic, you can draw a vertical line in the middle, going through the capacitors.
Conservation of charge means that charge on both sides of this line is constant.
Since Q=CV, then C1*V1 + C2*V2 will be constant, and that will allow you to calculate the capacitor voltages once steady state is reached.
The resistor does not influence the end result. A higher value will simply need more time to equalize voltage in your caps.
Now, the real amusing thing here is if you calculate the energy. Since E=1/2 CV^2 you'll probably notice that the energy at the end is lower than the energy at the beginning, even if the resistor is zero...