Differential Opamp: How to cut or reduce current consumption from supply rails
Use a GPIO pin of my MCU to switch Vdd supply to the opamp

Bad idea, it is almost never a good idea to switch on/off ICs by removing their supply. That will only work properly if you make sure that all other pins of the IC are also kept low. If only one pin has a higher voltage then that pin will power the chip via the ESD protection diodes. For more extensive explanation, watch this video by Dave from the EEVBlog. In this video Dave demonstrates the effect using a microcontroller, the same principle applies to opamps as well.

Use large resistors

Good idea but watch out that the input biasing current of the opamp is small enough. The LM358 is OK-ish in this respect with a typical input bias current of 10 nA so across a 1 M ohm resistor you'd get a voltage drop of 10 mV.
For a low power application with low currents, you already know that the LM358 is not a good choice, it has a typical quiescent supply current of 300 uA.
With an opamp using CMOS technology you can get much lower than that, also some opamps have an enable input that you could use to switch it off. But if you choose a very low power opamp that might not even be needed.
How to find such an opamp? Again Dave helps, showing us how to do a components search based on certain parameters in this video.