Isn't this very much related to the problem of a transversal in a Latin square? Suppose we have an $(n,n)$ bipartite graph with $n$ edge colors, such that every vertex has one edge of each color. This is equivalent to an $n\times n$ Latin square. A rainbow matching is a transversal of the Latin square. There is a conjecture (due to Ryser) that every Latin square with $n$ odd has a transversal, that is, a perfect rainbow matching. For even $n$, the conjecture (due to Brualdi) is that it has a partial transversal of length $n-1$ (i.e., a rainbow matching of cardinality $n-1$). To indulge in a little self-promotion, the best known result is that there exists a partial transversal of length $n -O(\log^2 n)$. There are also a number of results about transversals and partial transversals in near-Latin squares, which will probably be relevant to rainbow matching questions.

Isn't this very much related to the problem of a transversal in a Latin square? Suppose we have an $(n,n)$ bipartite graph with $n$ edge colors, such that every vertex has one edge of each color. This is equivalent to an $n\times n$ Latin square. A rainbow matching is a transversal of the Latin square. There is a conjecture (due to Ryser) that every Latin square with $n$ odd has a transversal, that is, a perfect rainbow matching. For even $n$, the conjecture (due to Brualdi) is that it has a partial transversal of length $n-1$ (i.e., a rainbow matching of cardinality $n-1$). To indulge in a little self-promotion, the best known result is that there exists a partial transversal of length $n -O(\log^2 n)$. There are also a number of results about transversals and partial transversals in near-Latin squares, which will probably be relevant to rainbow matching questions.

I guess the relevant Latin square question would be: does a random Latin square have a transversal with high probability? I know extensive calculations have been done which suggest that the answer is yes. I don't know whether anybody has proven this.