You may make a bet at any time; in particular, you may either make a bet when there are an odd number of remaining cards or an even number of remaining cards.
With this in mind, we will prove the claim that no strategy exists which can make the chances of winning greater than $50\%$, by considering $2$ cases. For each, we will look at $P(r,b)$ where $r$ and $b$ denote the number of red and black cards remaining, so that $P(r,b)$ is the probability that $r$ red cards remain and $b$ black cards. Obviously, the chances of winning are greater than $50\%$ if $P(r>b) > P(b>r)$.
Case $1$: The number of remaining cards is odd (say, $2n+1$).
Observe that $P(0,2n+1) = P(2n+1,0)$. [It's just as likely to have no black cards as it is to have no red cards]. Also $P(1,2n) = P(2n,1)$, and so on until (and including) $P(n,n+1) = P(n+1,n)$. In total, $P(r>b) = P(b>r)$, so the odds of winning are not greater than $50\%$.
Case $2$: The number of remaining cards is even (say, $2n+2$).
Using reasoning similar to Case $1$, we have $P(0,2n+2) = P(2n+2,0); P(1,2n+1) = P(2n+1, 1)$; and so on until (and including) $P(n,n+2) = P(n+2,n)$. For all these possible outcomes, $P(r>b) = P(b>r)$, so the chance of winning is not greater than $50\%$.
The only distinction in this case is the outcome where $(r,b) = (n+1,n+1)$. Sadly this does not help our chances. Since the number of cards remaining is equal for each suit, our probability of winning is still $50\%$.