Low quiescent linear Voltage Regulator

You don't want a switcher. Your input-output difference is so low (for most of the discharge curve the voltage is less than 3.8 V) that even at 90 % efficiency it won't do better much than an LDO.

Have a look at the Seiko S1167:

• Ground current typical 9 µA
• Available in 100 mV steps from 1.5 V to 5.5 V
• 1 % output voltage accuracy
• 150 mA output current
• 150 mV drop-out at 100 mA
• Shutdown input
• SOT23-5 package

edit
Found an even better one in the S1313:

• Ground current 0.9 µA
• 200 mA output current

Seiko doesn't give data for ground current under load, but in my experience you should count on 1 % of load current, so at 1 mA that would be around 10 µA, most likely less than 50 µA.

Here is a low power LDO voltage regulator (TPS79733) that consumes less than 2uA quiescent current whilst delivering up to 10mA to keep your MCU ticking over when it's asleep.

It has a fixed output of 3.3V and I think the clincher would be to arrange for your TPS63061 switcher to generate maybe 50mV more output (3.35V) so that when it gets activated, it sources all the higher current when you are taking 400mA.

In effect the LDO would not source any current because it would see an output voltage across its terminals that forces it to "open circuit" its pass-transistor.

Thus you don't need to enable/disable the LDO regulator.

My question is how to handle this switch-over phase. The LDOs which I looked at have reverse current protection and may be put into standby/shutdown mode. Is this sufficient to not destroy the LDO when the switcher starts working?

Handling the switch-over phase - enable the switcher when the MCU powers up Standby/shutdown for LDO protecting the LDO - the LDO will always be enabled.