Does a boosting circuit reduce battery voltage without any load?
Does boosting module play a role in reducing the battery voltage when no load connected?

What you're asking as called the quiescent current consumption of the boost converter.
This is the current that the boost converter takes when it outputs the boosted voltage ( 8 V) but no current is pulled (by the load) from that 8 V.
When I want to know this \$I_q\$ I look at the datasheet of the chip, on page 3 there's a table with the information:

The Shutdown current is the lowest but then the converter is shut-down so there will not be 8 V at the output!
There are two Quiescent Current figures, one for PFM and one for PWM.
Usually PFM is used when the load uses a small current, the converter then operates in small bursts and shuts down in between bursts. This saves supply current as we can see in the numbers: \$I_{q,PFM}\$ = 100 uA typically
PWM is used when the load is higher. The converter switches continuously. The \$I_{q,PWM}\$ has a higher value (1.6 mA typically) compared to the PFM mode but that usually is not an issue as the load will draw a lot of current as well. So the 1.6 mA that the converter needs is usually small compared to the load current.
So yes, the converter will take some current, it will take around 100 uA continuously. If that will drain your battery overnight depends on the capacity of your battery. With a 1000 mAh battery, at 100 uA it takes 10 thousand hours to discharge, that's 400 days. My guess is that you will be using a battery with more capacity than 1000 mAh so I do not think the converter can drain your battery overnight.
The change from 3.7 V to 3.67 V doesn't mean much, that can be due to temperature changes as well. I would measure the actual quiescent current, and if it is more than 200 uA (the specified maximum) then I would investigate what's going on.