Your series resistor is the current controlling device, but you made a calculation error. The resistor value is not 6 V divided by the current, but the voltage across the resistor divided by the current, and that's only 2.7 V, being 6 V minus the 3.3 V of the LEDs.
If you would now take a 150 Ω resistor, for (6 V - 3.3 V)/ 150 Ω = 18 mA, that 18 mA will be the total for all LEDs. So each LED would get only 3 mA. But don't decrease the resistor. Instead use one 150 Ω resistor per LED. Paralleling LEDs is a Bad Practice™; if their voltage is not exactly the same one LED will draw much more current than the other. Giving each LED its own series resistor balances the currents.
This may not be a solution for the problem that they go out, but it's something you'll have to fix anyway. Then check if the behavior changes.
edit
Someone who called himself Jack Goff (he's gone now, but soon will be back under yet another name) suggested the problem was thermal runaway, but that wasn't the case here.
A diode's forward voltage has a negative temperature coefficient, so as temperature rises the voltage drop decreases. Thermal runaway occurs if the decrease would cause a higher increase in current so that their product (power) increases, so that temperature increases. The forward voltage decreases further, current increases, etc.
JG suggested to "add series drop for each diode to cause V drop to increase rather than decrease for increasing current". Well, OP had such a device, it's called a resistor. That it was only one for all LEDs is OK to avoid thermal runaway. Firstly, the resistance was 330 Ω, so the total of all LED currents could never have exceeded 18 mA, and even then their voltage would have to go to zero. One LED will have the lowest voltage and therefore take most of the current. Even taking all of it it would only be 9 mA at 3 V, for a maximum dissipation of 27 mW (due to the maximum power transfer theorem). That's 40 % of the power at nominal current and voltage. A further voltage decrease will decrease power, and therefore temperature. So the resistor stabilizes the LED, no thermal runaway, but like I said the single resistor may cause unequal current distribution and therefore unequal brightness.
