I'll answer by analogy to give a clearer picture of what's going on. In a sense, a capacitor is like a storage tank for electrons. This means that a capacitor with a larger capacitance can store more charge than a capacitor with smaller capacitance, for a fixed voltage across the capacitor leads.
The voltage across a capacitor leads is very analogous to water pressure in a pipe, as higher voltage leads to a higher flow rate of electrons (electric current) in a wire for a given electrical resistance, per Ohm's Law.
Given the tank analogy, if I increase the diameter of a tank of water, but keep the starting level of water the same as it initially was, I have more water in the tank. When I poke a hole at the bottom of the tank shell, water will flow out and the tank level will drop. For both the small tank (initial conditions) and the large tank, the initial flow rate of water will be the same for a given hole size because the pressure at the bottom of the tank only depends on the height of the water above the hole, but it is obvious that the small tank of water will run dry first because there is less water in the tank. This means that the time constant of the small tank is smaller than it is for the large tank.
If I decrease the hole size (increase the resistance to flow), the time constant for both tanks will increase, but the small tank will always run dry first if both tanks start at the same level.
Regarding the title of this query, the rate of discharge of a capacitor is normally seen to be the rate at which charge is leaving the capacitor plates. This is the current in the associated circuit. How fast the voltage across capacitor plates is decreasing, and how fast the current in the associated circuit is decreasing, is related to the time constant of the circuit, which is NOT the current flowing in the circuit. In other words be careful not to confuse current in the circuit with the time constant of the circuit.