How does the CMRR of a fully differential amplifier change with input impedance imbalance?
No, this would not work.  The Whitlock amplifier works by bootstrapping the common-mode input impedance so that it is very high, so that the voltage divider effect from the source impedances becomes negligible, and therefore so do the effects of any imbalance in those source impedances.   

Your fully-differential amplifier is essentially acting as a buffer amplifier, hiding the source and load impedances from each other, so all the Whitlock amplifier can do is make impedance imbalances of the diff amp negligible.  

The source imbalance will only see the diff amp's input impedance, and if that is not extremely high, it will cause a voltage imbalance coming out of this stage.