Shut down regulator during sleep
This particular regulator features a power save pin (PS), which will switch the working mode over from constantly regulating into a mode where only short bursts of current are provided to boost the voltage above the set voltage point.

If your components can handle the extra voltage, that might be worth a try. I'd add some more capacity to the output to increase the off-phase of the regulator to reduce the needed current even further.



So with a rough calculation:
\$\frac{(3.38V-3.3V)*100\mu F}{10\mu A} = 0.8s\$

This is the time, a 100µF output capacitor would need to drop from 3.38V down to 3.3V witha load of 10µA (neglecting voltage dependencies). And the regulator would be on only some 5µs, so it is basically off the whole time.

As mkeith mentioned it might be worth considering a LDO, something like a TPS72733 which will have a quiescent current of 7.9µA (there should be some around going even lower). They have a high efficiency for battery powered applications and are a lot cheaper than switchmode regulators. The efficiency is roughly 88-89% at 3.3V out and 3.7V in and 200mA load, but only about 50% if the load is as low as the quiescent current and shutdown is not usable (depends on how the chip implements it). However you will loose the ability to drain the battery below 3.3V + dropout voltage (130mV @ 200mA for that part) which will decrease the interval at which the batteries have to be recharged. But on the other hand the batteries will last more recharge cycles if not discharged so deeply.

It should be noted that batteries like a constant low current draw more than a spiky high current draw, so the LDO might increase the interval again - but with the batteries you are going to use, drawing 1A for some microseconds should be fine. Battery lifetime is quite complex and not easy to predict, so to really find out, you'd probably need a prototype with both regulators and see what happens.