No, it makes no difference.
Number the balls according to when they will be drawn out of the bag, so that whoever chooses $4$th gets ball number $4$, etc. Now suppose you have a strategy for deciding when to pick that gets red more often than not. Suppose for a particular ordering you decide to pick at time $r$, so you get the $r$th ball. You must do exactly the same for the ordering which is the same for the first $r-1$ balls, but then is reversed after that (because you have exactly the same information at the point you decide to act). Thus, if you follow the strategy your chance of picking a red ball doesn't change if, instead of getting the next ball when you pick, you always get the $100$th ball when you pick. But in that case it makes no difference what you do (since you always get the same ball), and the chance of the $100$th ball being red is $50\%$.
This is counterintuitive because it seems like you should be able to wait until there is one more red ball than blue left, for more than $50\%$. But the point is that this isn't guaranteed to happen. When it does, you gain a small extra chance of getting a red ball, but on the rare times it doesn't, you lose much more.
[For those who are familiar with the concept, what's really going on is that the probability of the next ball being red is a martingale, and the optional stopping theorem applies.]