Two people answered that the Petersen graph is a counterexample. They are correct, but it is more interesting to note that any cubic graph with $4k+2$ vertices is a counterexample. The contracted graph has $2k+1$ vertices an $4k+2$ edges. Each edge colour can appear on at most $k$ edges so at least 5 colors are needed.

Two people answered that the Petersen graph is a counterexample. They are correct, but it is more interesting to note that any cubic graph with $4k+2$ vertices is a counterexample. The contracted graph has $2k+1$ vertices an $4k+2$ edges. Each edge colour can appear on at most $k$ edges so at least 5 colors are needed.

For $4k$ vertices: Take any simple quartic graph with $2k$ vertices and stretch out each vertex into a new edge joining two vertices of degree 3. This is the reverse of the operation given. So every simple quartic graph wit $4k$ vertices can be produced by contracting a perfect matching in a cubic graph. Is every quartic graph 4-edge-colourable? I don't think so; actually I think it is NP-hard.