You don't say what your motivating application is. It would be easier to sample without replacement but maybe that is not an option. You also suggest that you might be willing to limit the maximum number of times the most frequent item occurs. Any assumptions on the model change what can be said. I am just going to assume that a puzzlemaster, unknown to us, has put $n$ colored balls in a bag according to some private scheme she made up. We sample $s$ times with replacement and must draw conclusions as best we can. I will assume that $n$ is known but very large and that $0.1 n \lt s \lt 0.9 n$ but that is just a vague thought in the back of my mind. You do seem to want $s \lt n$
I think that you don't find a good match with species richness because you are not asking the right questions. Do a thought experiment about various outcomes and what you can conclude.
Suppose all $s$ balls turn out to be green. What can you conclude? Most of the balls are green. You can say with $90\%$ confidence that the number of green balls is at least $\alpha n$ for some $\alpha(n,s)$ which I won't attempt to suggest that I personally know how to compute. Suppose that you do know and you announce "Of the $n=1000000$ balls, we sampled $s$, they were all green, so I am $95\%$ confident that at least $870,000$ are green. You get a surprise hint "good guess!, exactly $900000$ are green. Now you know for sure that $2 \le N \le 100001.$ You get a further hint that the answer is either $n=2$ or $N=10001$. Can you say anything about the chance that $N \gt 10?$ I say no. OK now suppose you got just the first hint or even no hints at all. Does that make it easier to decide if $N \gt 10?$ How could it? And with no hints how confident can you be that there are any balls which are not green? I don't see why you would be.
NOW suppose that of your $s$ sampled they were green and red in the ratio 17 to 9. Then you can be $90\% $ confident that at least $\alpha n$ are red or green for the same $\alpha $ as before. But as far as saying if there are any other colors and if so how many, things are no better and no worse than before. What you can say with good confidence is that there are about twice as many green as red..
SO I think you can answer these questions (which you don't seem that interested in): Estimate how many balls might be a different color than any we have seen. For each color seen, estimate how many balls are this color.