That is the best way to do it. A couple things I would note on this. One of them, is the use of an empty string in the replace function. I know on small applications, that it doesn't make a serious difference, but an empty string uses memory space (well, more than a null value). A Character holds (I believe it's) 8 bytes. A Null Value holds like 1, or 2. So, in really big VB applications, changing "" to vbnullstring can speed up the app (benchmarks show by 40%) by using vbnullstring instead of "". So, you could redo the function as SearchString = Replace(SearchString, Chr(13), vbnullstring) to perform the same action, and save RAM usage. Again, I know this isn't extremely important, especially with modern machines, but it's good to know, and a nice practice to have.
The other thing, and I'm not sure if this is the intended result or not, but the above code will remove ALL chr(13's) from the string, not just the trailing one (when reading files, it's usually not a problem, since input will separate lines by newlines as it's read in, but a lot of apps that I have built require reading a file (or from a socket) 1 character at a time). An alternative method, if you are just wanting to remove the trailing newline, would be to use the left function SearchString = left(SearchString, Len(SearchString) -1).
I'm sure that the replace function is a fine solution for what you are doing, but I wanted to make sure to note the couple of exceptions.