Although this is a little late and the conversation has evolved away from this, I'm putting this out there to say...
...you're exactly right, and we do have the info...
Calculating Battery Drain Current
More than just the inefficiencies, there is also an increased amp draw from the battery to boost the voltage. If you are boosting a 3.7v source to 6v the circuit does this by drawing more amps from the source than it is applying to the load. What we don't have is any info on how much more current is pulled from the battery to boost voltage, but if we understand there is no free lunch in the conversion then it must be more efficient to use the battery at its native voltage.
...you're exactly right, and we do have the info...
Calculating Battery Drain Current