It's not a question of quicker but rather how much longer a more efficient solution would allow.
so lets say you double the 3.7 to 7.4 and then use an LDO 5V VREG. Further, assume your circuit draws 100 mA. Ignore efficieny for a minute. Your doubler will need to draw 200 mA from the battery - giving you a battery life of 50 hrs. during this 50 hrs, you are wasting 2.4*.1 (.24 Watts) as heat. that gives you a 67% (5/7.4) efficiency. However, your doubler has some loss as well so in practice it will be lower than 67%. All this, however, presumes your battery gives out a constant 3.7V - it doesn't so the analysis is more complex but sufficient for now.
In the case where you use a boost regulator you should be able to get to around 80% or better efficiency. The efficiency difference will translate into a longer run time.
Now, if I were you, I'd try to figure out how to run off of the 3.7V battery directly. You would need to deal with the changing voltage but it would give the absolutely longest run time. worth the effort.