Capacitor Varying Charging Voltage
I was wondering what would happen if you took a 2.7v super-cap and
  began to charge it with 2.5V then while continuing to charge it the
  input voltage drops to 1V.


This question carries an assumption that has an error. Once the capacitor is charged to 2.5V, if you reduced the supply voltage, the capacitor would have charge taken away from it i.e. the action of lowering the terminal voltage "discharges" the capacitor.

Q = CV under all circumstances and assumign C is constant (not always a good assumption), if you lower the voltage you lower the charge stored in the device.

If you started with 1V and raised it to 2.5V then you are indeed charging the capacitor. Another formula derived from Q = CV is the formula for current into (or out of) a capacitor. Differentiate both sides with respect to time (assuming C is constant) and you get: -

\$\dfrac{dQ}{dt} = C\dfrac{dV}{dt}\$

And of course dQ/dt = current. This tells you that if you apply a raming voltage (1 volt per second) a 1 farad capacitor, you will cause a constant current into the cap of 1 amp.