The only convex functions majorized by $\sin$ are the functions $h(x) = c$, where $c \le -1$. Hence $(\operatorname{conv} \sin)(x) = -1$.
To see this, suppose $h \le \sin$, $h$ is convex but $h$ is not constant. Then for some $x_1<x_2$, we have $h(x_1) \neq h(x_2)$. Suppose $h(x_1) < h(x_2)$, let $l(x) = \frac{x-x_2}{x_1-x_2}h(x_1) + \frac{x-x_1}{x_2-x_1}h(x_2)$, $l$ is affine and $\lim_{x \to \infty} l(x) = \infty$. Choose $x > x_2$, and write $x_2$ as a convex combination of $x_1,x$. This gives $x_2 = \frac{x_2-x_1}{x-x_1}x+\frac{x-x_2}{x-x_1}x_1$, and hence $h(x_2) \leq \frac{x_2-x_1}{x-x_1}h(x)+\frac{x-x_2}{x-x_1}h(x_1)$. Rearranging gives $l(x) \le h(x)$, which is a contradiction, as $h$ is bounded above by $1$ (since it is majorized by $\sin$). Hence $h$ is constant. Since $h \le \sin$, and $\sin (-\frac{\pi}{2}) = -1$, it follows that the constant is $\le -1$.
The following is a change from my previous answer, thanks to @byk7 (see the comments below) for catching a problem with my previous answer.
Here is another way of seeing this. Given a convex set $C \subset \mathbb{R}^n \times \mathbb{R}$, define $L_C(x) = \inf \{ \mu | (x,\mu) \in C \}$. It is straightforward to see (Theorem 5.3 Rockafellar) that $L_C$ is a convex function.
Now note that $\operatorname{conv} \sin=L_{\operatorname{co} ( \operatorname{epi} \sin)}$.
Since $\operatorname{co} ( \operatorname{epi} \sin) = \mathbb{R} \times [-1,\infty)$, we have $h(x) = \inf \{ \alpha | (\alpha,x) \in \operatorname{co} ( \operatorname{epi} \sin) \} = -1$.
Your conjecture is almost correct. The convex envelope of $F$ is strictly decreasing on the (bounded) interval $[0, 1]$, and strictly increasing or identically zero on the (unbounded) interval $[1, \infty)$. $F$ need not even be continuous for this conclusion, and the strict convexity near $x=1$ is also not needed.
In the following, $\hat F$ denotes the (lower) convex envelope of $F$, that is
$$
\hat F(x) = \sup \{ h(x) \mid \text{$h: \operatorname{dom}(F) \to \Bbb R$ is convex}, h \le F \} \, .
$$
Then we have the following result for unbounded intervals:
Let $F: [a, \infty)\to \Bbb R$ be strictly increasing with $F(a) = 0$. Then $\hat F$ is identically zero or strictly increasing on $[a, \infty)$.
And for bounded intervals:
Let $I = [a, b]$ or $I = [a, b)$ and $F: I \to \Bbb R$ be strictly increasing with $F(a) = 0$. Then $\hat F$ is strictly increasing on $I$.
For a proof of the statement about unbounded intervals we distinguish two cases:
Case 1: $\liminf_{x \to \infty} F(x)/x = 0$. Then also $\liminf_{x \to \infty} G(x)/x = 0$, and that implies that $G$ is identically zero on $[a, \infty)$.
Case 2: $\liminf_{x \to \infty} F(x)/x > 0$. Then $F(x) > cx$ for some constant $c > 0$ and $x \ge x_1 > a$.
For $x_0 \in (a, x_1)$ set $m = \min(c, \frac{f(x_0)}{x_1 - x_0})$ and consider the function $h(x) = m(x-x_0)$. $h$ is convex with $h \le F$, so that $\hat F(x) \ge h(x) > 0$ on $(x_0, \infty)$.
Since $x_0$ can be arbitrarily close to $a$ it follows $\hat F(x) > 0$ on $(a, \infty)$. This implies that $\hat F$ is strictly increasing on $[a, \infty)$.
A similar reasoning as in the second case can be used to prove the statement about bounded intervals.
Best Answer
If $F:(a,b) \to \Bbb R$ is bounded below (so that the convex envelope exists) and differentiable then its convex envelope $\hat F$ is continuously differentiable, i.e. a $C^1$ function. (The continuity of $F'$ is not needed for this conclusion.)
More generally one can show that if $F:(a,b) \to \Bbb R$ is bounded below, and differentiable at a point $c \in (a, b)$ then $\hat F$ is also differentiable at $c$.
This implies the desired conclusion: If $F$ is differentiable on $(a, b)$ then the same holds for $\hat F$. The derivative of a convex function is (weakly) increasing and cannot have jump discontinuities (because of Darboux's theorem). It follows that the derivative of $\hat F$ is continuous (see also Continuity of derivative of convex function).
In the following I'll write $G = \hat F$ for the convex envelope of $F$ (to save some keystrokes, and because $\hat{F}'$ renders ugly in MathJax).
Assume that$F$ is differentiable at $c$, but $G$ is not differentiable at $c$. Then $G_-'(c) < G_+'(c)$ where $G_-'$ and $G_+'$ denote the left and right derivative of $G$. Define $$ h(x) = \begin{cases} G(c) + G_-'(c)(x-c) & \text{ if } a < x \le c \, ,\\ G(c) + G_+'(c)(x-c) & \text{ if } c \le x < b \, . \end{cases} $$ Then $F(x) \ge G(x) \ge h(x)$ for all $x$.
Next show that $F(c) = G(c)$. Otherwise for sufficiently small $\epsilon$ the function $h_\epsilon$ obtained by replacing $h$ with a straight line segment on $[c-\epsilon, c+\epsilon]$ still satisfies $F(x) \ge h_\epsilon(x)$ for all $x$, but $h_\epsilon(c) > G(c)$. This contradicts the maximality of $G$.
It follows that $$ F(x) \ge \begin{cases} F(c) + G_-'(c)(x-c) & \text{ if } a < x \le c \\ F(c) + G_+'(c)(x-c) & \text{ if } c \le x < b \end{cases} $$ and that implies $$ F_-'(c) \le G_-'(c) < G_+'(c) \le F_+'(c) $$ in contradiction to the assumption that $F$ is differentiable at $c$.