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.
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 affine functions 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 latter functions the inequality in question is straightforward to check. In fact, the constant factor $2$ in the denominator on the right-hand side of the inequality in the OP can be replaced by the better factor $\pi$.
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.