$\newcommand\Om\Omega\newcommand\R{\Bbb R}$The answer is no. 

Indeed, for real $k>0$, let 
$$G_k:=\{(x,y)\in\Bbb R^2\colon x>1,|y|<k\sqrt x\}. $$
For $(x,y)\in G_2$, let 
$$f_0(x,y):=y^2/x-1.$$
For all $(x,y)\in\R^2$, let 
$$f(x,y):=\max\big(0,\sup_{(u,v)\in G_2\setminus G_1}[f_0(u,v)+(\nabla f_0)(u,v)\cdot(x-u,y-v)]\big),$$
where $\cdot$ stands for the dot product. The function $f$ is convex, being the pointwise supremum of affine functions. Moreover, $f$ is real valued, because $f_0(u,v)$ and $(\nabla f_0)(u,v)=(-\frac{v^2}{u^2},\frac{2 v}{u})$ are bounded in $(u,v)\in G_2\setminus G_1$. 

In fact, for all $(x,y)\in\R^2$, 
$$
f(x,y)=\begin{cases}
 0 & \text{ if }x\geq h(y),
 \\
 y^2/x-1 & \text{ if }\sqrt{x}<| y| \leq 2 x ,\\
 -4 x+4 y-1 & \text{ if }y\geq \max \left(0,2 x,x+\frac{1}{4}\right) ,\\
 -4 x-4 y-1 & \text{ if }-y\geq \max \left(0,2 x,x+\frac{1}{4}\right),
\end{cases}
\tag{1}\label{1}
$$
where 
$$
h(y):=\begin{cases}
 y^2 &\text{ if } | y| \geq 1/2, \\
| y| -1/4 & \text{ if }| y| <1/2.
\end{cases}
$$
In particular, $f=0$ on the nonempty open set $\Om=G_1$. Moreover, it is easy to see that the function $f_0$ is nonnegative and convex on $G_2$. So, $f=f_0$ on $G_2\setminus G_1$ and hence for $y=\frac32\,\sqrt x$ and $x>1$ one has 
$$(\nabla f)(x,y)=(\nabla f_0)(x,y)=\Big(-\frac9{4x},\frac3{\sqrt x}\Big)\to(0,0)$$
as $x\to\infty$, whereas $f(x,y)=f_0(x,y)=9/4-1\not\to0$. 

(The function $f$ may be not differentiable at points $(x,y)\in\R^2$ at which at least one of the inequalities in\eqref{1} turns into the equality. However, clearly the differentiability of $f$ has nothing to do with the essence of this problem. In particular,
one can smooth $f$ everywhere by (say) convolving it with a mollifier, while retaining the essential properties of $f$ noted above.) 

---

For an illustration, here is the graph $\{(x,y,f(x,y))\colon-1\le x\le10,|y|\le3\}$:

[![enter image description here][1]][1]


  [1]: https://i.sstatic.net/xF4SQc4i.png