Note that
\begin{equation}
\int_0^1\int_0^1 h(|x-y|)(f(x)-f(y))^2\,dx\,dy-2\int_0^1 h(x)dx=2D_f(h),
\end{equation}
where
\begin{equation}
D_f(h):=\int_0^1 dx \int_x^1 dy\,h(y-x)(f(x)-f(y))^2-\int_0^1 h.
\end{equation}
So, it is enough to find, for each $h$ as in the OP, a function $f\in\mathcal H$ such that $D_f(h)<0$.

We shall do a bit more than this -- by letting $h$ be any continuous function on $[0,1]$ such that
\begin{equation}
h(0)>h(1).
\end{equation}

Take any $t\in(0,1/2)$, and for all $x\in[0,1]$ let
\begin{equation}
f(x):=f_t(x):=\tfrac1{\sqrt{2t}}(I\{x<t\}-I\{x>1-t\}),
\end{equation}
where $I$ is the indicator. Then $f\in\mathcal H$.
Also,
\begin{equation}
2t(f(x)-f(y))^2=I\{x<t\}I\{y\ge t\}+I\{x\le1-t\}I\{y>1-t\}+2I\{x<t\}I\{y\ge1-t\}
\end{equation}
if $0<x<y<1$, whence, letting $t\downarrow0$, we have

\begin{equation}
\int_0^1 dx \int_x^1 dy\, h(y-x)(f(x)-f(y))^2=\frac{J_1+J_2+2J_3}{2t},
\end{equation}
where
\begin{align}
J_1&:=\int_0^t dx \int_t^1 dy\, h(y-x) \\
&= \int_0^t dx \int_{t-x}^{1-x} du\, h(u) \\
&= t\int_0^1 h- \int_0^t dx \Big(\int_0^{t-x}h+ \int_{1-x}^1 h\Big) \\
&= t\int_0^1 h- \int_0^t dx \Big(\int_0^{t-x}[h(0)+o(1)]+ \int_{1-x}^1 [h(1)+o(1)]\Big) \\
&= t\int_0^1 h- \frac{t^2}2\,[h(0)+h(1)+o(1)];
\end{align}
similarly,
\begin{align}
J_2&:=\int_{1-t}^1 dy\,\int_0^{1-t} dx\, h(y-x) \\
&= t\int_0^1 h- \frac{t^2}2\,[h(0)+h(1)+o(1)];
\end{align}
and
\begin{align}
J_3&:=\int_0^t dx \int_{1-t}^1 dy\, h(y-x) \\
&= t^2[h(1)+o(1)].
\end{align}
Collecting all the pieces, we have
\begin{equation}
D_f(h)=\frac{J_1+J_2+2J_3}{2t}-\int_0^1 h=[h(1)-h(0)+o(1)]t/2<0
\end{equation}
for small enough $t>0$ (depending on $h$).
Thus indeed, for each continuous function $h$ on $[0,1]$ such that $h(0)>h(1)$,
we have constructed a function $f\in\mathcal H$ such that $D_f(h)<0$.