By replacing $F(x)$ by $F(1-x)$ we may assume that $c\ge1/2$.
The problem is to determine a polynomial $G(x)=G_c(x)$ of minimal
possible degree, say $n$, such that $G(x)>0$ for $0 < x < 1$ and the derivative
of $F(x)=x(1-x)G(x)$ changes sign at $x=c$ and $F(x)\le F(c)$ for $0 \le x \le1$.
Clearly, $G_{1/2}(x)\equiv1$
and with a very little work $G_c(x)=x+c(2-3c)/(2c-1)$ for $1/2 < c\le2/3$, so $n=1$.
The case $n=2$ produces
$$G(x)=x^2+a\biggl(x-\frac{c(3c-2)}{2c-1}\biggr)-\frac{c^2(4c-3)}{2c-1},$$
so for $a \ge 0$ we have $G(x) > G(0)=c^2(3-4c)/(2c-1)$ on $x\in(0,1)$, while
the latter expression is non-negative for $2/3 < c\le3/4$. If $a <0$, then
either $G(x)$ is not positive for $0 < x < 1$ or $F(x)$ attains its maximum
at a different point of the interval $0 < x < 1$. This is however a little
bit technical to show. (For example, if we take
$a=1-3c$, then for the corresponding polynomial $G(x)$ we indeed have
$G(x)  > 0$, since
$$
G(x)\ge G\biggl(\frac{3c-1}2\biggr)=\frac{(2c+1)(c-1)^2}{4(2c-1)}.
$$
But $x=c$ is not the maximum of $F(x)=x(1-x)G(x)$ on the interval.)

If the above pattern remains, then for $n/(n+1) < c \le (n+1)/(n+2)$
the minimal possible degree of the polynomial $G(x)$ seems to be $n$
(so that $\deg F=n+2$), with the corresponding choice
$$
G_c(x)=x^n-\frac{c^n((n+2)c-(n+1))}{2c-1}.
$$
The limiting case $c=1$ is in favor of this observation: there is
no polynomial $F(x)\not\equiv0$ of the assumed form which attains
its maximum at $x=1$. So, the expected answer to the original
question would be $\deg F=\lceil 1/\min(c,1-c)\rceil$, where
$\lceil x\rceil=n$ when $n-1 <  x \le n$.