This is not quite an answer, but I'm going to post it anyway because I think it may lead to a real answer.
If we modify your question so that instead of $c_y$ we consider
$$d_y = d(y) = \inf\left\{z \in [y-h, y+h] : \int_{y-h}^{y+h} G(t)\,dt = 2 h G(z) \right\}$$ (note the closed interval instead of open), then we can show that the function $d$ is lower semicontinuous, and in particular is Lebesgue measurable. Fix a sequence $y_n \to y_0$ and set $d_0 = \liminf_{n \to \infty} d(y_n)$. We have to show $d(y_0) \le d_0$. Passing to a subsequence, we may assume $d_0 = \lim_{n \to \infty} d(y_n)$. Note that $y_n -h \le d(y_n) \le y_n + h$ by definition, so $y_0 - h \le d_0 \le y_0 + h$; in particular $d_0$ is finite.
Now for each $n$, by definition of $d$ there exists $z_n$ such that $d(y_n) \le z_n \le d(y_n) + \frac{1}{n}$ and
$$\int_{y_n-h}^{y_n+h} G(t)\,dt = 2 h G(z_n). \tag{*}$$
By the squeezing lemma, $z_n \to d_0$ and so by continuity of $G$ we have $2hG(z_n) \to 2hG(d_0)$. On the other hand, by the dominated convergence theorem we have
$$\int_{y_n-h}^{y_n+h} G(t)\,dt \to \int_{y_0-h}^{y_0 + h} G(t)\,dt.$$
(To see this, set $F_n = 1_{[y_n - h, y_n + h]}$ and note that $F_n(t) \to F_0(t)$ for all $t \ne y_0 \pm h$; in particular $F_n \to F_0$ almost everywhere. So $F_n G \to F_0 G$ almost everywhere, and is dominated by $G$ on some bounded interval large enough to contain all $[y_n - h, y_n + h]$.)
So when we pass to the limit in (*), we obtain
$$\int_{y_0 - h}^{y_0 + h} G(t)\,dt = 2 h G(d_0).$$
Therefore, $d(y_0) \le d_0$.
If we work with $c(y)$ instead, most of the argument above still works. But the very last line could fail because we might have $c_0 = y_0 - h$ in which case we cannot conclude $c(y_0) \le c_0$. I think there should be a way to fix this, but it's too late at night here to keep thinking about it...
No.
Your first hypothesis is true for any Lebesgue measurable function $f$; there is a sequence of continuous functions $f_n$ such that $f_n \to f$ almost everywhere. This is standard.
But there do exist Lebesgue measurable (even Borel) functions which are not a.e. equal to any function of Baire class 1.
For instance, it's a fairly standard exercise to construct a Borel set $B \subset [0,1]$ such that for any nontrivial interval $I$, we have $0 < m(B \cap I) < m(I)$. See this Math.SE post for details. Take $f = 1_B$.
Suppose $g = f$ a.e.; let $C_1 = g^{-1}(\{1\})$ and $C_0 = g^{-1}(\{0\})$, so $m(C_1 \triangle B) = 0$ and $m(C_0 \triangle B^c) =0$. As such, for any interval $I$ we have $m(C_i \cap I) > 0$. So in particular, on any interval, there are points at which $g=0$ and points at which $g=1$. Hence $g$ is nowhere continuous and thus not Baire class 1.
See also https://math.stackexchange.com/questions/15088/is-every-lebesgue-measurable-function-on-mathbbr-the-pointwise-limit-of-con. It's mentioned there that every Lebesgue measurable function is a.e. equal to some function of Baire class 2.
Best Answer
Let $U$ be an open and dense subset of $\mathbb{R}$ with finite measure. Let $g: \mathbb{R} \to \mathbb{R}^{\ge 0}$ be a continuous function with $\{g = 0\} = U^c$. Then define $f: \mathbb{R} \to \mathbb{R}$ by $f(x) = g(0)+\int_0^x g(t)dt$. Then $f \in C^1$, and $f' \equiv g$ means $f$ is strictly increasing (since any interval $(x,y)$ contains an interval lying in $U$ on which $g$ is strictly positive). So we have a strictly increasing $C^1$ function with zero derivative everywhere except for a finite measure set.