This can be done if $X$ has $\mathbb Q$-factorial singularities (but this is not a necessary condition!): Let $H$ be a relative ample effective divisor (not a bundle, divisor!). Then $f_*H$ is a Weil divisor on $X$ and if $X$ has $\mathbb Q$-factorial singularities, then some multiple of $f_*H$ is Cartier. Replacing $H$ with the appropriate multiple we may assume that $f_*H$ is Cartier. Then it can be pulled back and $f^*(f_*H)$ is relative relatively numerically trivial, hence $H-f^*f_*H$ is still relatively ample. Notice that $f^*f_*H-H$ is an effective exceptional divisor, hence trivial when restricted to the complement of the exceptional set.
Also notice that indeed if $f$ is a small contraction, then one cannot find such a divisor, but also, if $X$ has $\mathbb Q$-factorial singularities, then the exceptional set of $f$ has pure codimension $1$.
This can be done if $X$ has $\mathbb Q$-factorial singularities (but this is not a necessary condition!): Let $H$ be a relative ample effective divisor (not a bundle, divisor!). Then $f_*H$ is a Weil divisor on $X$ and if $X$ has $\mathbb Q$-factorial singularities, then some multiple of $f_*H$ is Cartier. Replacing $H$ with the appropriate multiple we may assume that $f_*H$ is Cartier. Then it can be pulled back and $f^*(f_*H)$ is relative numerically trivial, hence $H-f^*f_*H$ is still relatively ample. Notice that $f^*f_*H-H$ is an effective exceptional divisor, hence trivial when restricted to the complement of the exceptional set.
Also notice that indeed if $f$ is a small contraction, then one cannot find such a divisor, but also, if $X$ has $\mathbb Q$-factorial singularities, then the exceptional set of $f$ has pure codimension $1$.