A capacitor does have a "resistance"; but since a capacitor is fundamentally different than a resistor, it is not considered this way.
A resistor has a static resistance. It doesn't matter at what time it is measured, or what voltage is applied - the resistance stays the same.
A capacitor has a static capacitance. It DOES matter at what time it is measured, AND what voltage is applied - as it's "resistance" will be different!
The moment the switch is thrown, the capacitor seems like a short circuit (low resistance) because there is no charge on it's plates. How can "no charge" make large currents flow? Because of the fact that there is no charge yet, this imposes the flow of electrons. It is like an empty battery with zero internal resistance - if it is empty, then it will absorb every single bit of energy that can be put into it. So initially, a capacitor seems like a short or low resistance value, until it starts charging.
As the capacitor charges, it starts to behave less like a short. So one could say that it's "resistance" starts increasing (as an analogy.) Up to the point where it is completely full and refuses to take any more electricity - then it would seem like a very high resistance.
But note this is assuming the voltage is constant. If a capacitor is "charged" to say, 5v, then the voltage is suddenly changed to 10v, then the capacitor will react in exactly the same way as it did for the transition from 0v to 5v. (Initially a "short", then gradually behaving less so.) This is where Sixto's answer is spot-on - the rate-of-change determines the current, which is proportional. Instant voltage change = instant current change.
Now another interesting part is, this "stored charge on it's plates" is potential energy, meaning it can be extracted and used elsewhere. So for instance, charging a small capacitor to 3v, then placing a white LED across it's terminals, will cause the capacitor to discharge it's stored charge in reverse - through the LED - causing it to light for a short time.
The length of time it can power the LED is directly related to it's capacitance value: \$C=\frac{Q}{V}\$ The larger the capacitor physically (the more Q potential), the more capacitance, and thus, the more ability to absorb and release electrons for any given voltage.
Ohm's law always applies to DC - always - that's why it's called a law. But this is not DC... charge varies with time, volts vary, amps vary... so this is the AC domain.