Setting $\delta R =0$ on boundary of hypersurface Does requiring $\delta R=0$ on the boundary of hyper-surface create any restrictions or problems in deriving the field equations from Einstein-Hilbert Action? 
 A: I suggest you to take a look at Appendix E of Wald's General relativity book. There he derives all boundary terms which appear in the variation of Hilbert's action. There are only 3 terms coming from this variation. Two of them give Einstein's equation. The surface term comes from the other term, $g^{ab}\delta R_{ab}$, which is a total derivative, as explained by Wald. By applying Stokes theorem, we see that the boundary term only depends on derivatives of $\delta g_{ab}$, but one of them is parallel to the boundary. So, if you set "Dirichlet" boundary conditions $\delta g_{ab}=0$, this parallel derivative is zero and there is only one term left (giving the boundary term commented by Prahar above). However, if you choose to put all covariant derivatives of the variation to zero at the boundary, that is $\nabla_a\delta g_{bc} = 0$, you get no boundary term. Finally, if you set only parallel derivatives to zero, this condition is equivalent to putting $\delta g_{ab} = 0$, giving the same boundary term. Restricting second-order derivatives in GR is not a well-posed boundary condition (as for any hyperbolic evolution system).  
You can also check in that same appendix the Hamiltonian formulation of GR. The boundary term appears again in the Hamiltonian. This term depends only on canonical conjugate momentum $\pi_{ab}$, which in turns depends only on first derivatives of the hypersurface metric $h_{ab}$. 
I hope this helps.
