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, then 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 is straightforward to check.
Detail: The mentioned mixture representation of the convex function $v$ is as follows: for $x\in(-1,1)$, $$\begin{aligned} &v(x) \\ &=v(-1+)+\int_{[-1,x)}dz\,v'(z) \\ &=v(-1+)+\int_{[-1,1)}dz\,1(z<x)\,v'(z) \\ &=v(-1+)+\int_{[-1,1)}dz\,1(z<x)\,\Big(v'(-1)+\int_{[-1,z]}\,dv'(a)\Big) \\ &=v(-1+)+v'(-1)(x+1) \\ &\qquad\qquad+\int_{[-1,1)}dz\,1(z<x)\,\int_{[-1,z]}\,dv'(a) \\ &=v(-1+)+v'(-1)(x+1) \\ &\qquad\qquad+\int_{[-1,1)}\,dv'(a)\int_{[-1,1)}dz\,1(a\le z<x)\, \\ &=[v(-1+)+v'(-1)]+v'(-1)x+\int_{[-1,1)}\,dv'(a)\,v_a(x), \end{aligned}$$ and the Lebesgue--Stieltjes measure $dv'$ for the nondecreasing function $v'$ is $\ge0$.