How do I calculate the capacitor value (Farad & Voltage) given the pulse current, pulse duration & maximum tolerable voltage drop?
Capacitor Discharge through Constant Current Source

From the link, you can get the equation for constant current through a capacitor. This allows you to assume a worst case spec of 2 Amps for the entire 600uS.

Assuming you can virtually fully charge the capacitor in the off time (15% duty cycle). 

\$v(t)= \frac{1}{C}I t + v(0)\$.

\$ C = \frac{It}{v(t) - v(t_0) }\$

Plugging your numbers in, without taking into account the 0.1 ohm ESR you should end up with 0.003F. Keep in mind current is negative since you're discharging it. With 0.1 ohm ESR taking out 200mV of voltage and assuming the supply voltage is only 4.5V, you'll only have 200mV left to play with. Plugging in 4.5V-4.3V you'll end up with needing 0.006F.

If you don't want the worst case for a resistive load and instead want to go with something more accurate to save on cap costs you could use the RC equation:
\$V(t) = V_0(e^{-t/ RC}) \$
With that, you would have to integrate it and then solve for C. For R, you would use 4.5V/2Amps which you sort of implied was your maximum current.

Also, the original question asks what voltage the capacitor should be. Good design should be rated 25-50% above what's expected. So it should be rated at ~5.6V to 6.75V.