Storing power from Raspberry Pi 5 V then amping it up to 12 V at 1A
This is pretty straightforward. 

You use a boost converter with a nominal 12 volt output which is also current limited to an input current which will fit your power budget from the USB. Let's say your USB power supply can spare 100 mA.

Now let's look at your energy storage capacitor. Because you want it to supply about 12 volts for 20 seconds, and the capacitor voltage will drop under load, how big a capacitor do you need? Let's say, just as a starting point, that you'll accept a 2 volt drop in 20 seconds, or 0.1 volt/sec. During this time, assume that the charging current is insignificant.

The pertinent equation relates capacitance, current and voltage by $$\frac{dV}{dt}= \frac{i}{C}$$ In this particular case, we can talk about the voltage and time discrete differences rather than the instantaneous values, so $$\frac{{\Delta}V}{{\Delta}t}= \frac{i}{C}$$ or $$C = \frac{i{\Delta}t}{{\Delta}V} $$ Plugging in your values, i = 1.5, delta V = 2, delta t = 20, we get C = 15. That is 15 farads, not microfarads.

Now, as for charging. Assuming 100% efficiency, the charger must provide the same input power as output. The input power (current times voltage) must be 5 times 0.1, or 0.5 watts. 12 volts out means the output current must be (5/12) 0.1, or about 40 mA. Since the output current is 1.5 amps the charge time must be 1.5/.04 or about 37 times longer than the discharge time. In this case, that works out to about 12 1/2 minutes of charging time for each motor activation.

You can play games with this, of course. If you let the voltage drop from 12 to 8 volts, the capacitance can be half as great, so the charge time will be approximately halved.