No, an odd cycle is a counterexample. Similarly if there is a component which is an odd cycle.  But maybe these are the only cases?  (Haha, I forgot to click "save" and Mikhail beat me.)

Ok, wlog $G$ is connected. If all vertices have degree 2, it depends on odd or even length, as above. 

The cases where $G$ is a dumbbell or a theta-graph (two vertices with three paths joining them), or a collection of cycles with one common vertex, are easy to colour by *ad hoc* methods. So suppose $G$ is not one of those. Then there is a path $P$ of (possibly 0) degree 2 vertices between two distinct vertices of degree at least 3. Remove the edges and internal vertices of $P$ and apply induction.  Then put back $P$ and easily colour its edges as well.