$\newcommand{\al}{\alpha}
\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\varepsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}}
\newcommand{\PP}{\operatorname{\mathsf P}}
\newcommand{\EE}{\mathcal E}
\newcommand{\F}{\mathcal F}
\newcommand{\I}{\mathcal I}
\newcommand{\x}{\mathbf x}
\newcommand{\size}{\text{size}}
\newcommand{\pow}{\text{power}}
\newcommand{\st}{\text{stupid}}$
This conjecture is true. Of course, the hardest part of this problem is the presence of the $\min$. So, the crucial point in the proof is the following upper bound:
Lemma 1.
\begin{equation*}
\frac1{\min(m+q-1,x)}\le\frac Am+\frac Bx,
\end{equation*}
where
\begin{equation*}
A:=\frac{(x-q+1)^2}{(x+q-1)^2},\quad B:=1-A=\frac{4 (q-1) x}{(x+q-1)^2}, \tag{3}
\end{equation*}
\begin{equation*}
1<q\le x,\quad 1\le m\le x. \tag{4}
\end{equation*}
In what follows, (3) and (4) will always be assumed.
(Recall that the case $q=1$ was already verified by the OP, and it will also follow by continuity.) The proof of Lemma 1 is quite elementary, but a bit tedious, and omitted here.
By Lemma 1, the left-hand side (lhs) of the inequality in question is upper bounded by
\begin{equation*}
\sum_{m=1}^{x} m\,\Big(\frac Am+\frac Bx\Big) \binom{x}{m} \left( \frac{x-q}{x+q}\right)^m
=A (r-1)+\frac{B r (x-q)}{2 x},
\end{equation*}
where
\begin{equation*}
r:=\left(\frac{2x}{x+q}\right)^x.
\end{equation*}
So, the inequality in question reduces to
\begin{equation*}
A (r-1)+\frac{B r (x-q)}{2 x}\le
\frac{x-q+2}{x+q}\,(r-1),
\end{equation*}
which can be rewriten as
\begin{equation}
\de(q):=\de(q,x):=\ln r-\ln\frac{(x+1)^2-1-(q-1)^2}{2 x + 2 q-1}\ge0. \tag{5}
\end{equation}
It is straightforward but tedious to see that
\begin{multline*}
[(x+1)^2-1-(q-1)^2]\de'(q) \\
=2 q^3 (x+1)+q^2 \left(2 x^2+x-2\right)-2 q x \left(x^2+x-1\right)-x \left(2 x^3+x^2-4 x+1\right)<0
\end{multline*}
for
\begin{equation}
1\le q\le x-2/5. \tag{6}
\end{equation}
So, $\de(q)=\de(q,x)$ is decreasing in $q\in[1,x-2/5]$. Moreover,
\begin{multline*}
(5 x-1) (20 x-9) ( 120 x-49) \frac d{dx}\de(x-2/5,x) \\
=12000 x^3 \ln (5)-100 x^2 (24+127 \ln (5)) \\
+\left(12000 x^3-12700 x^2+4265 x-441\right) \ln
\left(\frac{x}{5 x-1}\right)\\
+5 x (512+853 \ln (5))-541-441 \ln (5).
\end{multline*}
It is straightforward but tedious to see that the latter expression is positive for all $x\ge1$. (One way to deal with such an expression is to notice that the derivative of a high enough order of an expression of the form $P(x)\ln R(x)$ is a ratio of polynomials if $P(x)$ is a polynomial and $R(x)$ is a ratio of polynomials.)
So, $\de(x-2/5,x)$ is increasing in $x\ge1$. Also, $\de(x-2/5,x)|_{x=2}=0.00187\ldots>0$. So, $\de(x-2/5,x)>0$ for $x\ge2$.
Recalling that $\de(q)=\de(q,x)$ is decreasing in $q\in[1,x-2/5]$, we see that $\de(q,x)>0$ for $x\ge2$ and $q\in[1,x-2/5]$.
Thus, we get the inequality in question for $x\ge2$ and $q\in[1,x-2/5]$. The case $x=1$ is trivial.
So, it remains to consider the case when $x\ge2$ and $q\in(x-2/5,x]$. This case is much easier than the one considered, because in this case the $\min$ does not cause trouble: indeed, for $q\in(x-2/5,x]$ and $m=2,\dots,x$ we have $\min(m+q-1,x)=x$, a constant. So, in this case the difference between the left- and right-hand sides of the inequality in question times $2 q x (x+q)$ equals $2 x (x^2 - q (x - 2)) - q ((x - q)^2 + 4 x)r$, with $r$ as before. Hence, the inequality in question can be rewritten here as
\begin{equation}
\de(q):=\de(q,x):=\ln r-\ln\frac{2 x (x^2 - q (x - 2))} {q r ((x - q)^2 + 4 x)}\ge0; \tag{7}
\end{equation}
here we use the same notation, $\de(q)=\de(q,x)$, for an expression (somewhat similar to but) different from the one in (5). Here, we have
\begin{multline*}
q (x + q) (x^2 - q (x - 2)) ((x - q)^2 + 4 x)\de'(q) \\
=q^3 (11 - 6 x) x^2 + x^4 (4 + x) + q x^3 (4 - 11 x - 2 x^2) \\
+
2 q^4 (2 - 3 x + x^2) + q^2 x^2 (-20 + 5 x + 6 x^2)<0
\end{multline*}
for all $q\in[1,x]$, so that $\de(q,x)$ is decreasing in $q\in[1,x]$.
Also, $\de(x,x)=0$ and hence $\de(q,x)>0$ for all $q\in[1,x]$.
Thus, the inequality in question is completely proved.
