Is there an additive time hierarchy theorem?
I don't see the need for your clarification, since through linear speed-up those sets are exactly the same. It is clear why you need to avoid using asymptotics, since $f(n)-g(n)$ and $f(n)$ are asymptotically the same, but not using asymptotics doesn't somehow make the linear speed-up theorem not true.

Some further insight can be gained. $g(n)$ is actually $g(f(n))$,i.e. it must depend on $f(n)$ to work for all functions. Why? Suppose g(n) is non-constant. Pick $f(n) = 2 g(n)$ . Then clearly $f(n)$ and $f(n)-g(n)$ are equal by the linear speed-up theorem. If $g(n)=c$ is constant, then we can pick a non-constant $f(n)$ so that the conjecture doesn't hold.

What if we allow dependency on $f(n)$? Well then by taking the known deterministic hierarchy theorem and doing some basic arithmetic we can see that any $g(f(n))$ that is $o((1 - \frac{1}{ \log f(n)} ) f(n) )$ works. However I am not sure if just by stating the same theorem in a slightly different form opens up the possibility of some new insight.