Is There a Complete Problem for the Class of Turing Decidable Problems?
Generally a class having a complete problem under a nice class of reductions
implies that the class can be enumerated. 
$\mathsf{R}$ is not computably enumerable, therefore it does not have a complete problem with respect a nice class of reductions.

Here is the argument:

Assume that there is a complete problem $A$ for $\mathsf{R}$.
Therefore for any problem in $\mathsf{R}$ can be obtained from 
a reduction (let's say polynomial time many-one reductions) combined with $A$.
We can computably enumerate the reductions, 
therefore we can computably enumerate $\mathsf{R}$.
But $\mathsf{R}$ is not computably enumerable
(otherwise we could diagonalize).

In the literature look for the set of total recursive/computable functions.