Yes, it can always be done, while having a $k$-regular simple graph at every step. This is a very old result, I think from the 1970s.[1] A later result of Richard Taylor is that if the first and last graphs are connected then you can arrange for all the intermediate graphs to be connected too.[2] RIchard also showed the same for 2-connectivity.[3] All these results hold for arbitrary degree sequences and not just regular graphs.
[1] To be added....
[2] Constrained switchings in graphs. Combinatorial mathematics, VIII (Geelong, 1980), pp. 314–336, Lecture Notes in Math., 884, Springer, Berlin-New York, 1981.
[3] Taylor, R. Switchings constrained to 2-connectivity in simple graphs. SIAM J. Algebraic Discrete Methods 3 (1982), no. 1, 114–121.