$\newcommand\R{\mathbb R}\newcommand\ep{\varepsilon}$Such a statement does not hold for any $n\ge2$. Indeed, in view of the natural embedding of $\R^2$ into $\R^n$ for $n\ge2$, without loss of generality $n=2$.
The inequality in question is $$2u^\top f(x)u^\top x\le\ep|u|^2+\frac C\ep\,(u^\top x)^2 \tag{1}\label{1}$$ for all $x,u$ in $\R^2$, where $|\cdot|$ is the Euclidean norm.
Suppose that this is true. Take any $t\in(0,1)$. For all $x=(x_1,x_2)\in\R^2$, let $f(x):=f_t(x):=y+tx$, where $y:=(-x_2,x_1)$. Then $f$ is $2$-Lipshitz.
For each $x\in\R^2$, let now $u=f(x)$. Then \eqref{1} becomes $$2t(1+t^2)|x|^4\le\ep(1+t^2)|x|^2+\frac C\ep\,t^2 |x|^4.$$ Letting here $|x|\to\infty$ and diving by $t$, we get $$2(1+t^2)\le\frac C\ep\,t$$ for all $t\in(0,1)$. Letting now $t\downarrow0$, we get $2\le0$. $\quad\Box$