Unfortunately, one can construct an $h(x)$ such that it is untrue that $\psi^{\prime } (x) >0$ for sufficiently small $x$. To begin with, separate the integration intervals into $y<x$ and $y>x$, and substitute $t=x-y$ or $t=y-x$ such that $h$ appears with argument $t$, $h(t)$ everywhere. Then, $\psi $ takes the form $$ \psi (x) = x + \frac{\int_{x}^{1-x} dt\, t\, h(t)}{2\int_{0}^{x} dt\, h(t) + \int_{x}^{1-x} dt\, h(t) } $$ and therefore, $$ \psi^{\prime } (x) = 1 - \left. \frac{1}{\left[ 2\int_{0}^{x} dt\, h(t) + \int_{x}^{1-x} dt\, h(t) \right]^{2} } \right[ \\ ( (1-x) h(1-x)+xh(x)) \left( 2\int_{0}^{x} dt\, h(t) + \int_{x}^{1-x} dt\, h(t) \right) + \left. (h(x)-h(1-x)) \int_{x}^{1-x} dt\, t\, h(t) \right] $$ So, for $\psi^{\prime } (x)$ to remain positive, the denominator must exceed the numerator. Consider $h(t)=(1/2)(1+\theta (x_0+\epsilon -t))-\delta t$, with an arbitrarily small slope $\delta $, and with the step function smeared out to continuously drop from $1$ to $0$ on the arbitrarily narrow interval $]x_0 , x_0 +2\epsilon [$. Evaluating at $x=x_0 $, we have, up to arbitrarily small corrections, $$ \int_{0}^{x_0 } dt\, h(t) = x_0 \ , \ \ \int_{x_0 }^{1-x_0 } dt\, h(t) = \frac{1}{2} -x_0 \ , \ \ \int_{x_0 }^{1-x_0 } dt\, t\,h(t) = \frac{1}{4} (1-2 x_0 ) $$ and so, $$ \psi^{\prime } (x=x_0 ) = 1-\frac{3/8 + x_0/2 + x_0^2 /2}{1/4 +x_0 +x_0^2 } $$ which is negative for sufficiently small $x_0 $.

Unfortunately, one can construct an $h(x)$ such that it is untrue that $\psi^{\prime } (x) >0$ for sufficiently small $x$. To begin with, in order to get a better sense of what the moving parts are here, separate the integration intervals into $y<x$ and $y>x$, and substitute $t=x-y$ or $t=y-x$ such that $h$ appears with argument $t$, $h(t)$ everywhere. Then, $\psi $ takes the form $$ \psi (x) = x + \frac{\int_{x}^{1-x} dt\, t\, h(t)}{2\int_{0}^{x} dt\, h(t) + \int_{x}^{1-x} dt\, h(t) } $$ and therefore, $$ \psi^{\prime } (x) = 1 - \left. \frac{1}{\left[ 2\int_{0}^{x} dt\, h(t) + \int_{x}^{1-x} dt\, h(t) \right]^{2} } \right[ \\ ( (1-x) h(1-x)+xh(x)) \left( 2\int_{0}^{x} dt\, h(t) + \int_{x}^{1-x} dt\, h(t) \right) + \left. (h(x)-h(1-x)) \int_{x}^{1-x} dt\, t\, h(t) \right] $$ So, for $\psi^{\prime } (x)$ to remain positive, the denominator must exceed the numerator. Consider $h(t)=(1/2)(1+\theta (x_0+\epsilon -t))-\kappa t$, with, in order to formally satisfy the conditions of the OP, an arbitrarily small downward slope $\kappa $, and with the step function smeared out to continuously drop from $1$ to $0$ on the arbitrarily narrow interval $]x_0 , x_0 +2\epsilon [$. Evaluating at $x=x_0 $, we have, up to arbitrarily small corrections, $$ \int_{0}^{x_0 } dt\, h(t) = x_0 $$ $$ \int_{x_0 }^{1-x_0 } dt\, h(t) = \frac{1}{2} -x_0 $$ $$ \int_{x_0 }^{1-x_0 } dt\, t\,h(t) = \frac{1}{4} (1-2 x_0 ) $$ and so, $$ \psi^{\prime } (x=x_0 ) = 1-\frac{3/8 + x_0/2 + x_0^2 /2}{1/4 +x_0 +x_0^2 } $$ which is negative for sufficiently small $x_0 $.

Unfortunately, one can construct an $h(x)$ such that it is untrue that $\psi^{\prime } (x) >0$ for sufficiently small $x$. To begin with, separate the integration intervals into $y<x$ and $y>x$, and substitute $t=x-y$ or $t=y-x$ such that $h$ appears with argument $t$, $h(t)$ everywhere. Then, $\psi $ takes the form $$ \psi (x) = x + \frac{\int_{x}^{1-x} dt\, t\, h(t)}{2\int_{0}^{x} dt\, h(t) + \int_{x}^{1-x} dt\, h(t) } $$ and therefore, $$ \psi^{\prime } (x) = 1 - \left. \frac{1}{\left[ 2\int_{0}^{x} dt\, h(t) + \int_{x}^{1-x} dt\, h(t) \right]^{2} } \right[ \\ ( (1-x) h(1-x)+xh(x)) \left( 2\int_{0}^{x} dt\, h(t) + \int_{x}^{1-x} dt\, h(t) \right) + \left. (h(x)-h(1-x)) \int_{x}^{1-x} dt\, t\, h(t) \right] $$ So, for $\psi^{\prime } (x)$ to remain positive, the denominator must exceed the numerator. Consider $h(t)=(1/2)(1+\theta (x_0+\epsilon -t))$, with the step function smeared out to continuously drop from $1$ to $0$ on the arbitrarily narrow interval $]x_0 , x_0 +2\epsilon [$. Evaluating at $x=x_0 $, we have, up to arbitrarily small corrections, $$ \int_{0}^{x_0 } dt\, h(t) = x_0 \ , \ \ \int_{x_0 }^{1-x_0 } dt\, h(t) = \frac{1}{2} -x_0 \ , \ \ \int_{x_0 }^{1-x_0 } dt\, t\,h(t) = \frac{1}{4} (1-2 x_0 ) $$ and so, $$ \psi^{\prime } (x=x_0 ) = 1-\frac{3/8 + x_0/2 + x_0^2 /2}{1/4 +x_0 +x_0^2 } $$ which is negative for sufficiently small $x_0 $.

Unfortunately, one can construct an $h(x)$ such that it is untrue that $\psi^{\prime } (x) >0$ for sufficiently small $x$. To begin with, separate the integration intervals into $y<x$ and $y>x$, and substitute $t=x-y$ or $t=y-x$ such that $h$ appears with argument $t$, $h(t)$ everywhere. Then, $\psi $ takes the form $$ \psi (x) = x + \frac{\int_{x}^{1-x} dt\, t\, h(t)}{2\int_{0}^{x} dt\, h(t) + \int_{x}^{1-x} dt\, h(t) } $$ and therefore, $$ \psi^{\prime } (x) = 1 - \left. \frac{1}{\left[ 2\int_{0}^{x} dt\, h(t) + \int_{x}^{1-x} dt\, h(t) \right]^{2} } \right[ \\ ( (1-x) h(1-x)+xh(x)) \left( 2\int_{0}^{x} dt\, h(t) + \int_{x}^{1-x} dt\, h(t) \right) + \left. (h(x)-h(1-x)) \int_{x}^{1-x} dt\, t\, h(t) \right] $$ So, for $\psi^{\prime } (x)$ to remain positive, the denominator must exceed the numerator. Consider $h(t)=(1/2)(1+\theta (x_0+\epsilon -t))-\delta t$, with an arbitrarily small slope $\delta $, and with the step function smeared out to continuously drop from $1$ to $0$ on the arbitrarily narrow interval $]x_0 , x_0 +2\epsilon [$. Evaluating at $x=x_0 $, we have, up to arbitrarily small corrections, $$ \int_{0}^{x_0 } dt\, h(t) = x_0 \ , \ \ \int_{x_0 }^{1-x_0 } dt\, h(t) = \frac{1}{2} -x_0 \ , \ \ \int_{x_0 }^{1-x_0 } dt\, t\,h(t) = \frac{1}{4} (1-2 x_0 ) $$ and so, $$ \psi^{\prime } (x=x_0 ) = 1-\frac{3/8 + x_0/2 + x_0^2 /2}{1/4 +x_0 +x_0^2 } $$ which is negative for sufficiently small $x_0 $.