Yes: Just take $u(x,y):=v(x)$. 

Indeed, one can use Green's formula to show this, as is done in [Christian Remling's answer][1]. 

More generally, the result holds for any convex function $v$, without requiring it to be in $C^2$ -- of course, if $u$ is not required to be in $C^2$ and if the derivative $v'(x)$ is understood as (say) the right derivative of $v$ at $x$ for $x\in[-1,1)$. Indeed, then the convex function $v$ is a mixture of the constant functions, the identity function, 
and the functions of the form $v_a$ with $a\in[-1,1)$, where $v_a(x):=\max(0,x-a)$, and for any one of these functions the inequality in question is straightforward to check. 

  [1]: https://mathoverflow.net/a/463229/36721