There is no absolute guarantee. You can have an arbitrarily large multiple and the decode can still fail when a set of entries exist that form a cycle, it just becomes quite unlikely as the overhead goes up.

One of the ways of hybridizing iblt and exact algebraic techniques like the minisketch library I link in my other post is to staple a small algebraic sketch to the iblt. If the iblt is successful you're done, if it gets stuck you use take the recovered elements out of the algebraic sketch and decode that. It's fast to decode the algebraic sketch in spite its O(n^2) behavior because it's small, and it'll always be successful if there are few enough elements (unlike the iblt).

Sadly this still doesn't give a guarantee since you might have more elements in a cycle than the size of the backup, but small cycles are more likely than big ones so there exists a range of sizes where it's more communications efficient than a larger iblt.