capacitance calculations
Use C = I*T/deltaV

Assuming that the relay holds in until 9 V and starting from 12 V, you need
C = 20m*8/(12-9) = 53 mF. These are not readily available or practical. It is possible to get capacitors this large (supercaps), but they are generally only rated at 2.5 V. 

You would be better to use a solid-state switch (MOSFET): start with a higher supply V (e.g. 40 V would allow you to use 20m*8/(40-9) = 5000 uF which is nearly reasonable (although quite large -- about the size of a soda can).