For even $n$, I claim that nobody has a winning strategy, and therefore both players have drawing strategies. 

To see this, observe first that by the fundamental theorem of finite games, we know that either one of the players has a winning strategy, or both players have drawing strategies. 

Next, I claim that Bob has a drawing strategy, which is simply to use the copying idea of user Mohemnist in the comments. He should simply pair up opposite vertices and make the coloring anti-symmetric. In this way, any string of red vertices is matched by a symmetric string of blue vertices, and therefore Bob can ensure that he will not lose. 

But now, it follows that Bob cannot have a winning strategy, since Alice can pretend to be Bob by a strategy-stealing argument. Basically, this amounts to the observation that it cannot be advantageous to go second. That is, Alice can simply start by coloring any vertex red, and thereafter pretend to be the second player, following the winning strategy for Bob, but with swapped colors. If that strategy should ever direct her to color the already-colored vertex, then she can simply take another free move. 

So we've shown that Bob has a drawing strategy and cannot have a winning strategy. It follows that Alice cannot have a winning strategy, and so we must be in the case of the fundamental theorem where both players have drawing  strategies.