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 $x\in[-1,1)$ (and the left derivative for $x=1$). 

Indeed, with $u(x,y):=v(x)$ for all $x$ and $y$, the inequality in question is invariant with respect to taking positive/conical mixtures of functions $v$. That is, if the inequality in question holds for each suitably integrable function $v_b$ (in place of $v$) with $b$ in a set $B$, then the inequality in question holds for the positive mixture (used in place of $v$) of the form $\int_B\mu(db)\,v_b$ of the $v_b$'s, where $\mu$ is any nonnegative measure over $B$ such that the integral $\int_B\mu(db)\,v_b$ exists in a suitable sense. 

It remains to note that the convex function $v$ is a positive mixture of the constant functions, the identity function $\text{id}$, its opposite $-\text{id}$, 
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 (in fact, with the constant factor $\pi$ in the numerator on the right-hand side of the inequality, better than the corresponding constant factor $2$ in the OP) is straightforward to check. 

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**Detail:** The mentioned mixture representation of the convex function $v$ is as follows: for $x\in(-1,1)$, 
$$\begin{aligned}
&v(x) \\ 
&=v(-1+)+\int_{[-1,x)}dz\,v'(z) \\ 
&=v(-1+)+\int_{[-1,1)}dz\,1(z<x)\,v'(z) \\ 
&=v(-1+)+\int_{[-1,1)}dz\,1(z<x)\,\Big(v'(-1)+\int_{[-1,z]}\,dv'(a)\Big) \\ 
&=v(-1+)+v'(-1)(x+1) \\ 
&\qquad\qquad+\int_{[-1,1)}dz\,1(z<x)\,\int_{[-1,z]}\,dv'(a) \\ 
&=v(-1+)+v'(-1)(x+1) \\ 
&\qquad\qquad+\int_{[-1,1)}\,dv'(a)\int_{[-1,1)}dz\,1(a\le z<x)\, \\ 
&=[v(-1+)+v'(-1)]+v'(-1)x+\int_{[-1,1)}\,dv'(a)\,v_a(x) \\  
&=[v(-1+)+v'(-1)]+v'(-1)x+\int_{[-1,1)}\,\mu_{v'}(da)\,v_a(x) \\  
\end{aligned}$$
where $\mu_{v'}$ is the (nonnegative) Lebesgue--Stieltjes measure corresponding to the nondecreasing function $v'$. So, 
$$v=C+A\,\text{id}+\int_{[-1,1)}\,\mu_{v'}(da)\,v_a,  $$
where $C:=C_v:=v(-1+)+v'(-1)$ and $A:=A_v:=v'(-1)$ are real numbers. 

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