there are functions $f(n)$ and $g(n)$ such that $g(n) = o(1)$ and $f(n-g(n)) \neq \Theta((f(n))$
Thought about $f(n) = |sin(n)|,\ g(n)= \frac1n$ then $f(n-g(n))= |sin(n-\frac1n)|$ and then for any $n_0\geq0,\ c>0$ there is an $n>n_0$ but $f(n-g(n)) > c\cdot f(n)$ so for $n=2\pi n_0$ will get $ |sin(2\pi n_0-\frac1{\pi n_0)})|> c\cdot|sin(2\pi n_0)| = 0$
but i think $f$ should be a positive function $>0$.
need to prove\disprove with that definition.
With a hint from Steven Stadnicki, an example was found: $g(n)=1/n$, $f(n)=2^{n^3}$. Indeed, $$f(n-g(n))=2^{(n-1/n)^3} = 2^{n^3-3n+3/n-1/n^3} =\Theta(2^{n^3-3n})=o(2^{n^3})$$