What will happen if a capacitor is given a very large amount of charge within a very small period of time? When a capacitor reaches its maximum potential, and charge is still being given to it, it just leaks through. 
A teacher asked us to consider it similar to trying to pour a jug of water into a relatively smaller glass. When the glass is full, the water just overflows.
But when you suddenly flip over a jug trying to fill the glass in the smallest time possible you end up actually spilling more water and the glass doesn't even get filled up to the top. Does anything similar happen to a capacitor when an insane amount of charge is suddenly passed through it? 
What would even happen in such a scenario?
 A: 
When a capacitor reaches its maximum potential, and charge is still being given to it, it just leaks through.

This isn't generally true. In fact it's not true for any kind of capacitor I'm familiar with.
The more typical behavior if you over-charge a capacitor is that the electric field between the plates becomes too strong for the dielectric material, and an arc or spark is created, which damages the dielectric and plates. This could result in the plates being shorted together permanently, or the capacitance being reduced. Often these changes are permanent, and the capacitor is no longer suitable for whatever function it had in the circuit it was part of.

Does anything similar happen to a capacitor when an insane amount of charge is suddenly passed through it?

Capacitors generally have an equivalent series resistance (ESR). This can partly come from the resistance of the wires, and of the plates. Or it can model other loss mechanisms in the dielectric material. In any case, just like a real resistor, running current through this equivalent resistance heats up the capacitor. If you heat the capacitor up too much, it can burn out or otherwise be damaged (melt the dielectric, for example). 
This happens more often when too much AC current is passed through the capacitor (even if the voltage limits aren't violated), but a single too-fast charging or discharging of the capacitor can also cause over-heating if the capacitor ESR is too high relative to its heat dissipation capability.
A: In case of ideal capacitor, we have known equation for charging:
$$
U(t)=U_{max} \cdot \left( 1-e^{t/\tau}\right)
$$
To understand your problem we have to thinh, from where this equation comes from. If you are familiar with calculus we can determine differential of charge $d e$. If you arrent, you can understand $de$ as a werry small change of charge ($de=\Delta e; \Delta e \rightarrow 0$). How much charge will in small amount of time go on the plate of capacitor depends on voltage on the capacitor, resistor, trough which we are charging (or internal resistance of capacitor) and charging voltage. 
$$
I=\frac{de}{dt}=\frac{U_{charging}-U_{capacitor}}{R}
$$
In order to charge capacitor, there must be a voltage difference between capacitor and charging voltage. So when the capacitor is fully charged, we cant just add some charge, becouse the current (wich transports charge) won't flow if there is no voltage difference. 
So in order to increase charge, we have to increase voltage. In ideal case this works just fine, and capacitor countines charging to higher voltage. But in real life you can observe, that capacitors are marked with their maximal voltage. 
Betwen two plates of capacitor, there is normally an insulator, which is never an ideal non-conducting substance. At some point, the electric field in the insulator $\textbf{E}$ will, be so strong, that current will start flowing trough it. This phenomena depends on material, and cant be really simply explained. It depends on 
crystal structure of insulator for instance. Voltage, when this happens is called breakdown voltage. Normally material is deformed, when this happens. So in everyday case, if you charge capacitor with to high voltage, it will explode, because current will start flowing trough insulator weary rapidly and vapourize it. 
I can give you analogy with a dam and water. A pipe is pouring water into an ideal dam, which is so high, that it can't be overfilled. But the weight of water can break it. When there is to high voltage on your capacitor, it breaks the dam, and water starts flowing (if the capacitor isn't fully destroyed).
