Yes, indeed. Suppose $\mu,\nu$ are Borel probability measures on the metric space $(\Omega,\rho)$. The metric need not arise from a norm. Let $K(\mu,\nu)$ denote the space of couplings of $\mu,\nu$, i.e. Borel probability measures on $\Omega \times \Omega$ that project to $\mu$ in the first coordinate and to $\nu$ in the second coordinate. Recall that
$$W(\mu,\nu)=\inf_{\lambda \in K(\mu,\nu)} \, \, \Bigl\{ \int_{\Omega \times \Omega} \rho(x,y) \,d\lambda(x,y)\Bigr\} \,.$$
Suppose that $$\forall x,y \in \Omega \qquad |f(x)-f(y)|\le L\rho(x,y)^\alpha \,,$$
where $0<\alpha<1$.
Then for any $\lambda \in K(\mu,\nu)$, we have
\begin{eqnarray} \Bigl|\int_{\Omega} f \, (d\mu-d\nu) \Bigr|& \le&\int |f(x)-f(y)| \, d\lambda(x,y) \le L \int _{\Omega \times \Omega} \rho(x,y)^\alpha \, d\lambda(x,y) \\ &\le&
L \Bigl(\int _{\Omega \times \Omega} \rho(x,y) \, d\lambda(x,y)\Bigr)^\alpha \tag{*}\,,
\end{eqnarray}
where the last inequality is an application of Holder's inequality [1] for the functions $(x,y) \mapsto \rho(x,y)^\alpha$ and the constant $1$, with exponents $p=1/\alpha$ and $q=1/(1-\alpha)$.
Alternatively, the last inequality in $(*)$ can be obtained from an application of Jensen's inequality [2] for the convex function $t \mapsto t^{1/\alpha}$ on $[0,\infty)$.
Taking infimum over $\lambda \in K(\mu,\nu)$ in $(*)$ gives
$$\int_\Omega f(x)(d\mu - d\nu) \le LW(\mu,\nu)^\alpha \,,$$
as required.
[1] https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality
[2] https://en.wikipedia.org/wiki/Jensen%27s_inequality#Measure-theoretic_and_probabilistic_form
