Here's a sketch of a proof of a stronger statement: the coefficients of the Taylor series for $\log{}_2F_1(a,b;a+b+c;x)$ are rational functions of $a$, $b$, and $c$ with positive coefficients.

To see this we first note that
$$\begin{aligned}
\frac{d\ }{dx} \log {}_2F_1(a,b;a+b+c;x) &=
\frac{\displaystyle
\frac{d\ }{dx}\,{}_2F_1(a,b;a+b+c;x)}{{}_2F_1(a,b;a+b+c;x)}\\[3pt]
&=\frac{ab}{a+b+c}\frac{{}_2F_1(a+1,b+1;a+b+c+1;x)}{{}_2F_1(a,b;a+b+c;x)}.
\end{aligned}
$$
Then
$$
\begin{gathered}
\frac{{}_2F_1(a+1,b+1;a+b+c+1;x)}{{}_2F_1(a,b;a+b+c;x)}
= \frac{{}_2F_1(a+1,b+1;a+b+c+1;x)}{{}_2F_1(a,b+1;a+b+c;x)} \\
\hfill\times
\frac{{}_2F_1(a,b+1;a+b+c;x)}{{}_2F_1(a,b;a+b+c;x)}.\quad
\end{gathered}
$$
We have continued fractions for the two quotients on the right.
Let $S(x; a_1, a_2, a_3, \dots)$ denote the continued fraction
$$\cfrac{1}{1-\cfrac{a_1x}
{1-\cfrac{a_2x}
{1-\cfrac{a_3x}
{1-\ddots}
}}}
$$
Then
$$\begin{gathered}\frac{{}_2F_1(a+1,b+1;a+b+c+1;x)}{{}_2F_1(a,b+1;a+b+c;x)}
= S \left( x;{\frac { \left( b+1 \right) \left( b+c \right) }{ \left( a
+b+c+1 \right) \left( a+b+c \right) }},
\right.\hfill\\
\left.
{\frac { \left( a+1 \right)
\left( a+c \right) }{ \left( a+b+c+2 \right) \left( a+b+c+1 \right)
}},
{\frac { \left( b+2 \right) \left( b+c+1 \right) }{ \left( a+b+c+3
\right) \left( a+b+c+2 \right) }},
\right.\\
\hfill
\left.
{\frac { \left( a+2 \right)
\left( a+c+1 \right) }{ \left( a+b+c+4 \right) \left( a+b+c+3
\right) }},\dots \right)
\end{gathered}
$$
and
$$\begin{gathered}
\frac{{}_2F_1(a,b+1;a+b+c;x)}{{}_2F_1(a,b;a+b+c;x)}
=S \left( x,{\frac {a}{a+b+c}},
{\frac { \left( b+1 \right) \left( b+c
\right) }{ \left( a+b+c+1 \right) \left( a+b+c \right) }},
\right.\hfill\\
\left.
{\frac {
\left( a+1 \right) \left( a+c \right) }{ \left( a+b+c+2 \right)
\left( a+b+c+1 \right) }},
{\frac { \left( b+2 \right) \left( b+c+1
\right) }{ \left( a+b+c+3 \right) \left( a+b+c+2 \right) }},
\right.\\
\hfill
\left.
{\frac {
\left( a+2 \right) \left( a+c+1 \right) }{ \left( a+b+c+4 \right)
\left( a+b+c+3 \right) }}, \dots\right)
\end{gathered}
$$
The first of these continued fractions is Gauss's well-known continued fraction, and the second can easily be derived from the first. It follows from these formulas that the coefficients of the Taylor series for $\log{}_2F_1(a,b;a+b+c;x)$ are rational functions of $a$, $b$, and $c$ with positive coefficients.