A counterexample is given by $$ f(x,y)=(x^4-y^3)^2+y . $$ Clearly, $f(R, R^{4/3})=R^{4/3}\ll R^2 = \left( \min \{ R, R^{4/3} \} \right)^2$, so $f$ doesn't satisfy your condition.

However, $f$ is coercive: If $y\le 0$, then $f(x,y)\ge x^8+y^6 +y \ge x^8 + (1/2)y^6$, say, for large $|(x,y)|$. If $y\ge 0$, then $f(x,y)\ge \max\{ (x^4-y^3)^2, y\} \to\infty$ as $|(x,y)|\to\infty$, $y\ge 0$.

A counterexample is given by $$ f(x,y)=(x^4-y^3)^2+y . $$ Clearly, $f(R, R^{4/3})=R^{4/3}\ll R^2 = \left( \min \{ R, R^{4/3} \} \right)^2$, so $f$ doesn't satisfy your condition.

However, $f$ is coercive: If $y\le 0$, then $f(x,y)\ge x^8+y^6 +y$ and now either $|y|$ is not large and there are no problems or if $|y|\gg 1$, then $y^6+y\ge (1/2)y^6$, say. 

If $y\ge 0$, then $f(x,y)\ge \max\{ (x^4-y^3)^2, y\} \to\infty$ as $|(x,y)|\to\infty$, $y\ge 0$.