Let $p$ and $q$ be probability functions, say on $\mathbb{R}$. Suppose that we know $ |p-q|\leq \epsilon $ for some $\epsilon>0$ on $\mathbb{R}$. Let $X\sim p$ and $\tilde{X}\sim q$ be random variables that follow distributions defined by $p$ and $q$.
The answer here https://mathoverflow.net/questions/384911/perturbative-approach-starting-from-a-probability-distribution-approximated-form shows
$$|\mathbb{E}[f(X)]-\mathbb{E}[f(\tilde{X}))]|\leq C\epsilon$$ for $f$ such that $\int_{\mathbb{R}}|f|<\infty$.
Do we have a path-wise estimate $$\mathbb{E}[|f(X)-f(\tilde{X})|^n]\leq \text{small}$$ for $n$ suitable, e.g. $n=2$?
Even for $f(x)=x$, if $X$ and $\tilde{X}$ are independent we have the density of $X-\tilde{X}$ via convolution, but the independence might not be the case (indeed we expect $X$ and $\tilde{X}$ are pretty much the same).
Let $X$ have uniform distribution on $(-1,1)$ and $Y=-X$. Let $f$ be any integrable function such that $f(x)=x$ for $|x| \leq 1$. Then you are asking if $E|2X|^{n}$ is small. It is clearly not small for any $n$. [Here $p=q$!].