Do the math instead of asking us. (300 µA)(500 ms)/(100 µF) = 1.5 V. Can the microcontroller and whatever else it needs to do the shutdown still run from 1.8 V. If no, then your 100 uF cap is clearly inadequate.
However, instead of trying to hold up a regulated supply after the regulator, it is usually better to hold up the supply into the regulator. That way the voltage the micro is running from will be constant during the shutdown period. For example, if the 3.3 V supply is made by a buck regulator from the 32 V supply, then a diode followed by the holdup cap on the input of this buck regulator would do it.
You need (3.3 V)(300 µA)(500 ms) = 500 µJ. Lets say you need twice that on the input of the buck switcher to be conservative, so 1 mJ. Let's say the buck switcher needs 5 V minimum. To see where we're at, let's see how much energy a simple 10 µF ceramic cap can provide dropping from 32 V to 5 V. (32 V)²(10 µF)/2 = 5.12 mJ and (5 V)²(10 µF)/2 = 125 µJ, so the energy available from a 10 µF cap is 5 mJ. That's 5 times more than we conservatively estimated is required. In theory that means a 2 µF cap would do it, but ceramic caps are usually ±20%, and you're not going to save anything by using 2 µF instead of 10 µF. Note that it needs to be good to 35 V.
Scld made a good point in a comment that should be mentioned here. The calculations above assume the voltage drops from 32 V to 5 V during the backup power period. However, this only starts after the input voltage has already dropped some. You need to do the energy calculation starting with whatever voltage threshold you are using to detect that input power has dropped out. The energy in a capacitor is proportional to the square of the voltage, so at 1/2 voltage there is only 1/4 the energy available.
It looks like the 10 µF in the example above would still work for reasonable threshold votlages. Let's say you use 24 V as the threshold to detect input power lost. (24 V)²(10 µF)/2 = 2.9 mJ, leaving about 2.8 mJ available down to 5 V. That's still a good margin above the 1 mJ we decided was needed on the input of the buck regulator, so that's fine. Note however, that this also shows 10 µF is not enough if the power-out detection threshold is significantly lower. Keep in mind that the starting voltage you use for these caclulations is the lowest is could be worst case with all tolerances taken into account, and after the diode from the main 32 V supply.
Another point is that there would almost certainly be at least 10 µF on the input of the switcher anyway. If you set the power-out detection threshold appropriately, you may not need to add any parts at all.