Is there a sufficient condition for a regular graph to have a 1factorization (i.e. being able to pack all of its edges into disjoint perfect matchings, and excluding one vertex if the number of vertices is odd)?
Being a complete graph on an even number of vertices is a sufficient condition. This being said, testing whether a 3regular graph is 3edgecolorable is NPhard ( http://dx.doi.org/10.1016/j.ipl.2008.05.015 ) You can have certificates that a graph requires more than 3 colors by computing its fractional chromatic index (feasible in polynomial time with a LP solver and a maximum weighted matching algorithm at hand). This is checking whether your graph contains an overfull subgraph ( http://en.wikipedia.org/wiki/Overfull_graph ) For this, you can have a look at : Daniel Ullman and Edward Scheinerman  Fractional Graph Theory http://www.ams.jhu.edu/~ers/fgt/fgt.pdf Nathann 


Perhaps you already know this, but every bipartite regular graph is 1factorable (see e.g. these lecture notes). 

