The critical case is the linear cases $f(t) = \alpha t$, and $H = \alpha^{-1}F$. I will show that if asymptotically $f'(t)$ is bounded below (so if $f$ is asymptotically superlinear) than the desired conclusion hold.

The first step is to prove that the conclusion holds if $H$ is eventually $< F$, using merely a convexity argument.

Suppose there exists some $t_0$ such that for every $t > t_0$ we have that $H(t) < F(t)$. Notice that $\dot{H} = F$ and $\ddot{H} = \dot{f} F > 0$ we have that $H$ is convex. We are interested in the quantity

$$ H^{-1}(s) - F^{-1}(s) $$

which is positive for all sufficiently large $s$ when $H$ is eventually $< F$.

By convexity, we have that for any $r < s$

$$ H^{-1}(s) < H^{-1}(r) + (H^{-1})'(r) \cdot (s-r) $$

Since $H < F$ we can shoose $r = H\circ F^{-1}(s)$ and get

$$ H^{-1}(s) < F^{-1}(s) + (H^{-1})'\circ H \circ F^{-1}(s) \cdot (s - H\circ F^{-1}(s)) $$

Now

$$ (H^{-1})' = \frac{1}{H'\circ H^{-1}} $$

so we get

$$ H^{-1}(s) < F^{-1}(s) + \frac{1}{s} \cdot (s - H\circ F^{-1}(s)) $$

So in particular, you have that $H^{-1}(s) - F^{-1}(s)$ is bounded, since $H\circ F^{-1}$ is increasing.

Using that $F^{-1}$ grows unboundedly this implies

$$ H^{-1} - F^{-1} = o(F^{-1})$$

as desired.

Notice that in this argument I have not explicitly used the conditions $f' = \omega(t^{-1})$ and $\log(f') = o(f)$; however, the condition that $H(t) < F(t)$ can be derived from strengthened versions of your assumption.

For example, a sufficient condition to guarantee that $H(t) < F(t)$ eventually for every solution $H$ is that $f'(t) > 1 + \epsilon$ for all sufficiently large $t$.

This can be generalized further. Suppose that $f'(t)$ is lower-bounded by $\delta > 0$.

Define the function $\tilde{f}(\tau) = f(2 \delta^{-1} \tau)$. Define $\tilde{F}$ and $\tilde{H}$ analogously. We have that $\tilde{F} = F(2 \delta^{-1}\tau)$ and $\tilde{H} = \frac{\delta}{2} H(2\delta^{-1}\tau)$

Applying the above argument we get

$$ H^{-1}(2 \delta^{-1} s) - F^{-1}(s) < \frac{2}{\delta} - \frac{1}{s} H\circ F^{-1}(s) $$

Now

$$ F^{-1}(s) = f^{-1} \ln(s) $$

so

$$ F^{-1}(2 \delta^{-1} s) = f^{-1} ( \ln(s) + \ln 2 - \ln \delta) $$

Using that $f'$ is bounded below by $\delta$, we have that

$$ |F^{-1}(2 \delta^{-1} s) - F^{-1}(s) | < \delta \ln(2 \delta^{-1}) $$

is also bounded, so that we conclude

$$ H^{-1}(s) - F^{-1}(s)$$

is bounded, and the unbounded growth of $F^{-1}$ takes care of the rest.

The remaining case is when $\liminf_{t\to\infty} f'(t) = 0$, which is probably the case you are really interested in to boot, but unfortunately I don't yet see a proof for.

The simplest cases are the linear cases where $f(t) = \alpha t$, and $H = \alpha^{-1}F$. Modelling on those cases, I will show that if asymptotically $f'(t)$ is bounded below (so if $f$ is asymptotically superlinear) than the desired conclusion hold.

The first step is to prove that the conclusion holds if $H$ is eventually $< F$, using merely a convexity argument.

Suppose there exists some $t_0$ such that for every $t > t_0$ we have that $H(t) < F(t)$. Notice that $\dot{H} = F$ and $\ddot{H} = \dot{f} F > 0$ we have that $H$ is convex. We are interested in the quantity

$$ H^{-1}(s) - F^{-1}(s) $$

which is positive for all sufficiently large $s$ when $H$ is eventually $< F$.

By convexity, we have that for any $r < s$

$$ H^{-1}(s) < H^{-1}(r) + (H^{-1})'(r) \cdot (s-r) $$

Since $H < F$ we can shoose $r = H\circ F^{-1}(s)$ and get

$$ H^{-1}(s) < F^{-1}(s) + (H^{-1})'\circ H \circ F^{-1}(s) \cdot (s - H\circ F^{-1}(s)) $$

Now

$$ (H^{-1})' = \frac{1}{H'\circ H^{-1}} $$

so we get

$$ H^{-1}(s) < F^{-1}(s) + \frac{1}{s} \cdot (s - H\circ F^{-1}(s)) $$

So in particular, you have that $H^{-1}(s) - F^{-1}(s)$ is bounded, since $H\circ F^{-1}$ is increasing.

Using that $F^{-1}$ grows unboundedly this implies

$$ H^{-1} - F^{-1} = o(F^{-1})$$

as desired.

Notice that in this argument I have not explicitly used the conditions $f' = \omega(t^{-1})$ and $\log(f') = o(f)$; however, the condition that $H(t) < F(t)$ can be derived from strengthened versions of your assumption.

For example, a sufficient condition to guarantee that $H(t) < F(t)$ eventually for every solution $H$ is that $f'(t) > 1 + \epsilon$ for all sufficiently large $t$.

This can be generalized further. Suppose that $f'(t)$ is lower-bounded by $\delta > 0$.

Define the function $\tilde{f}(\tau) = f(2 \delta^{-1} \tau)$. Define $\tilde{F}$ and $\tilde{H}$ analogously. We have that $\tilde{F} = F(2 \delta^{-1}\tau)$ and $\tilde{H} = \frac{\delta}{2} H(2\delta^{-1}\tau)$

Applying the above argument we get

$$ H^{-1}(2 \delta^{-1} s) - F^{-1}(s) < \frac{2}{\delta} - \frac{1}{s} H\circ F^{-1}(s) $$

Now

$$ F^{-1}(s) = f^{-1} \ln(s) $$

so

$$ F^{-1}(2 \delta^{-1} s) = f^{-1} ( \ln(s) + \ln 2 - \ln \delta) $$

Using that $f'$ is bounded below by $\delta$, we have that

$$ |F^{-1}(2 \delta^{-1} s) - F^{-1}(s) | < \delta^{-1} \ln(2 \delta^{-1}) $$

is also bounded, so that we conclude

$$ H^{-1}(s) - F^{-1}(s)$$

is bounded, and the unbounded growth of $F^{-1}$ takes care of the rest.

The remaining case is when $\liminf_{t\to\infty} f'(t) = 0$, which is probably the case you are really interested in to boot, but unfortunately I don't yet see a proof for.