The energy in a capacitor is
$$ E = \frac{1}{2}CV^2 $$
So at 2.7 V in two capacitors, you would have
$$ E = 2\:\text{capacitors} \cdot \frac{1}{2} \cdot 360\:\mathrm F \cdot (2.7\:\mathrm V)^2 = 2624.4 \:\text{joules} $$
The amount of energy you are taking per second is
$$ 0.5\:\mathrm A \cdot 5\:\mathrm V = 2.5\:\mathrm{J/s} $$
So, in a perfect capacitor with a perfect power supply, you could run this for
$$ \frac{2624.4\:\mathrm J}{2.5\:\mathrm{J/s}}
= 1049.76\:\mathrm s
= 17.496 \:\text{minutes} $$
Note, that this is only with a perfect capacitor. Supercaps tend to have a high series resistance that loses energy. In addition, the regulator has an efficiency that will vary according to the input, likely going down in various ranges, and certainly not really going to 0 Volts. This means you will have to have the amount of energy to maintain the minimum voltage, and add to it the amount you actually consume for the time you want.
To run a 5 V power supply with 90% efficiency at 0.5 A for 2 hours, or 7200 seconds requires a specific amount of energy to the input of the power supply:
$$ 5 V \cdot 0.5 A \cdot 7200 s \cdot \frac{1}{0.90} = 20000 J$$
Note that the 90% efficiency effectively increases the amount of energy needed.
In addition, power supplies typically won't run down to 0 V. So there must be extra energy to handle the amount that is never discharged. We'll call the minimum voltage Vmin.
$$ E_{required} = E_{@Vmax} - E_{@Vmin} = \frac{CV_{max}^2}{2} - \frac{CV_{min}^2}{2}$$
$$ E_{required} = \frac{C}{2} \cdot (V_{max}^2 - V_{min}^2 )$$
So for a Vmax = 2.7 V and a Vmin = 0.5V
$$ 20000 J = \frac{C}{2} \cdot (7.04 V^2)$$
$$ C = \frac {20000 J \cdot 2}{7.04} = 5681.8 F$$
This could be one huge capacitor or many smaller adding up in parallel.
However, note that I haven't considered any losses due to series resistance. That just adds on the total capacitance needed, but by using multiple parallel capacitors, you tend to reduce the effective resistance as you are putting them in parallel.
The relevant formula for this sort of circuit is below:
$$C_{min} = \frac{t_{HOLD}\cdot I_{OUT}}{\Delta V}$$
In your case, neglecting the voltage drop of the current-limiting resistor, and assuming you can run down to 3.3V, this becomes
$$C_{min} = \frac{10\text{ s}\cdot I_{OUT}}{5\text{V} - V_{diode} - 3.3\text{V}}$$
Ballpark numbers:
Use a Schottky diode, and you'll maybe have 300 mV for the diode drop. Let's say the Pi draws 0.4A.
With these numbers, the equation above gives 2.9F. So your capacitor is close, but might not be quite enough.
A 12V battery, charger, and buck converter might be a better way to do this. You'd at least be able to avoid the giant capacitor.
Best Answer
From your parameters, your supercap would discharge in 1848 seconds to 1.8v under a constant 12mA draw.
$$Bt(seconds) = C(Vcapmax - Vcapmin) / Imax$$
If it's only active for 100ms every minute it has a duty cycle of:
$$100ms / 60000ms = 0.0016667%$$
It would last ~1.1 million minutes, or about two years. That is excluding the sleep mode draw however. At 20uA, interestingly enough your total active mode power consumption would be about the same as your total sleep mode power consumption, so we can easily estimate that including the sleep mode (which will be 99.84443% of the total time), your device will last for about a year from fully charged to 1.8v. You could extend this quite a bit by adding a high efficiency buck-boost, provided you don't add too many losses with it. Some modern boost converters can yield 1.8v out from as low as 0.25v in.