Comparison of two quadratic forms

Difference.
$$R-Q = (2x^2+2y^2)-(x^2+2xy+2y^2)=x^2-2xy.$$
Complete the square: $x^2-2xy=(x-y)^2-y^2$, which is not always $\ge0$. So the direct difference is not obviously positive. Instead:

Alternative. Compare with eigenvalues. $Q$ has matrix $A_Q=\begin{pmatrix}1&1\\1&2\end{pmatrix}$, eigenvalues $(3\pm\sqrt5)/2$. $R$ has matrix $2I$, eigenvalues $2$. The largest eigenvalue of $A_Q$ is $(3+\sqrt5)/2\approx2.618>2$, so $Q\le R$ is false in general.

Counterexample. At $(x,y)=(1,-1)$: $Q=1-2+2=1$ and $R=2+2=4$, so $Q<R$. At $(x,y)=(1,1)$: $Q=1+2+2=5$ and $R=4$, so $Q>R$.

Conclusion. The inequality $Q\le R$ is false in general. The largest eigenvalue of $Q$'s matrix exceeds $2$.