Where to start?
First, your HV unit is not a transformer. It is a high-voltage module. It puts out a pulsing high voltage.
So, the high voltage transformer was made to make arcs, and as such
does not allow ... a no-load situation.
You are correct. Your capacitor will store charge and will destroy your module. Furthermore, you have taken the input power specification, (and it would be nice if you would share with us where you got the numbers) of 18 watts at 4.5 volts, and then calculated an output power of 18 watts/20 kv, giving .9 mA. This is wrong. Input power does not equal output power.
You have also failed to notice the part of the description that reads
Work: Input voltage 1.5V ~ 3V can work for about 1 minute, the input
voltage exceeds 3V continuous work does not allow more than 30 seconds
In other words, you will not be able to run your unit for more than 30 seconds at a time. And it doesn't say how long you need to let the unit recover from 30 seconds operation, either.
Having failed to understand your HV unit, you have equally failed to understand your water capacitor. To begin with, ultrapure water is does not have an infinite resistivity. It is, in fact, 18 Mohm-cm. For your described capacitor, this amounts to ~ 2 Mohm. At 20 kV, that will require 10 mA which you cannot provide. You mention a "coating on the plates" which you believe will prevent current flow, but you do not describe it, and I hope you will forgive my suspicion that it may not work as you plan.
You speak of your desire not to break down the dielectric (the water) and then talk about how pulsing or AC voltages will avoid this. You seem to be unaware that the breakdown potential for pure water is ~3 MV/m. Since the spacing for your capacitor is on the order of 12.5 mm, the breakdown voltage will be ~ 37 kV, or twice your voltage, and you have no need to worry.
Given the dielectric constant of water (80), the effective capacitance of your capacitor will be ~ 100 pF. Since your module has a diode in its output, you are correct in thinking that you will get just a DC output, since there is no major discharge path. You cannot discharge your capacitor from the input side. Any such discharge mechanism would have to be synchronized with the HV pulses, and would have to withstand 20 kV. Such a circuit will be neither simple nor cheap.
Your selection of series resistor is very strange. Your calculation for voltage drop make it clear that you have used the input figure of 4.5 amps, and this has absolutely no application to the output.
So, overall, I have no way give you advice, except to suggest that you do something else. I don't see any instance where you seem to actually understand how your proposed circuit would work, or why, and that is not a good starting point.
Best Answer
If by charges you mean electric charges, then no, a capacitor does not store charges. This is a common misconception, maybe due to the multiple meanings of the word charge. When some charge goes in one terminal of a capacitor, an equal amount of charge leaves the other. So, the total charge in the capacitor is constant.
What capacitors store is energy. Specifically, they store it in an electric field. All the electrons are attracted to all the protons. At equilibrium, there are equal numbers of protons and electrons on each plate of the capacitor, and there is no stored energy, and no voltage across the capacitor.
But, if you connect the capacitor to something like a battery, then some of the electrons will be pulled away from one plate, and an equal number of electrons will be pushed on to the other plate. Now one plate has a net negative charge, and the other has a net positive charge. This results in a difference in electrical potential between the plates, and an increasingly strong electric field as more charges are separated.
The electric field exerts a force on the charges which attempts to return the capacitor back to equilibrium, with balanced charges on each plate. As long as the capacitor remains connected to the battery, this force is balanced by force of the battery, and the imbalance remains.
If the battery is removed, and we leave the circuit open, the charges can't move, so the charge imbalance remains. The field is still applying a force to the charges, but they can't move, like a ball at the top of a hill, or a spring held under tension. The energy stored in the capacitor remains.
If the capacitor terminals are connected with a resistor, then the charges can move, so there is a current. The energy that was stored in the capacitor is converted to heat in the resistor, the voltage decreases, the charges become less imbalanced, and the field weakens.
Further reading: CAPACITOR COMPLAINTS (1996 William J. Beaty)