If an NP complete problem 'A' is polynomial time reducible to another problem 'B' does that imply 'B' is also NP complete?
All that you know is that R is NP-hard.

To show that R is NP-complete, you need to show that it is in NP.  But that is not automatically true.

As an example, let S be the question "Is there a path of cost at most l that visits every node of a graph G?" and R be the question, "What is the cheapest path that visits every node of a graph G?"  These are variants of the Traveling Salesman problem.  S is in NP while R is believed not to be.  The issue being that given a proposed solution, you just add it up for S.  But for R, there is no known polynomial way to verify that you actually have the best path.