Say we throw $b$ blue balls and $r$ red balls uniformly into $n$ boxes. The probability that no box contains a red as well as a blue ball is then, by the inclusion exclusion principle:
$$p = \sum_{k,\ m} \binom{n}{k} \binom{n-k}{m} (-1)^{n-m-k} \left(\frac{m}{n}\right)^r \left(\frac{k}{n}\right)^b$$
I'm interested in upper bounding $p$. If we assume the red balls are thrown first, and without replacement, we can lower bound it as
$$\left(1-\frac{r}{n}\right)^b$$
Or 1 if both $r$ and $b$ are as big as $n$. Note $e^{-rb/n}$ is another good approximation, though neither a lower nor an upper bound.
Can you see any way to get a simple upper bound? I've tried to look into asymptotic expansions, but I don't see how I can be done.

Not what you are looking after, but a simple asymtoptic result would be: assume the number of blue (red) balls in each box are iid Poisson with $\lambda_B = b/n$ ($\lambda_R=r/n$), hence the probability that no box has red and blue balls would be
$$P \approx \left(1-(1-e^{-\lambda_B})(1-e^{-\lambda_R}) \right)^n=\left(e^{-b/n} + e^{-r/n} - e^{-(r+b)/n} \right)^n \tag{1}$$
I would bet that this is only useful for $n$ much bigger than $r,b$, and that it overestimates $P$, but it does not seem easy to prove that it's an upper bound.
Another approximation: the expected number of boxes with red balls is $E_r=n \left(1-(1-1/n)^r\right)$. Equating this (expected) number with the actual number, we get
$$P \approx \frac{(n-E_r)!(n-E_b)!}{n! \, (n-E_r-E_b)!} \tag{2}$$
A few results (
pscomes from simulation,pa1andpa2are the above approximations)