OK, here are some (unpleasant, but apparently correct) observations. We are still trying to solve $w^4+\dot w^2=H$ in real time. I'll also assume that you want $w\ge 0$, i.e., you do not want to change the direction of your rotation in the process. Note that allowing the direction change won't help much in the sense that the final conclusion will stay the same but the argument will get more involved.
Recall (see my AoPS post) that invariant quantities are $\dot H/H^{5/4}$ and $\ddot H/H^{3/2}$. Assume that we want to create a control satisfying $w^4+\dot w^2\in [(1-\delta)H,(1+\delta)H]$ for all reasonable regimes $H$. Suppose that the pilot is not allowed to change $H$ or $\dot H$ abruptly but he can change $\ddot H$ any time and in any way within some reasonable limits, so $\ddot H$ does not need to be continuous. I believe, this is more or less what happens in reality but if I am wrong here, correct me.
Observation 1. Ascending regimes are asymptotically stable.
What it means is that if $H$ is any ascending regime that starts at some distant past time moment $-T$ and equals some constant $H_0$ up to moment $0$ (we shall call such regimes canonical), then any non-negative solution of the equation $w^4+\dot w^2=H$ that is defined on the entire interval $[-T,T]$ has essentially the same values $w(t),\dot w(t)$ as the canonical solution whose initial condition is given by $w=H_0^{1/4}$ on $[-T,0]$.
Observation 2. Suppose that $H$ is a canonical ascending regime and $w$ is the corresponding canonical solution with $\dot w\ge 0$. Suppose also that $w_1$ satisfies $w_1^4+\dot w_1^2<H$ on $[-T,T]$. Then $w_1\le w$. Indeed, we clearly have $w_1\le w$ on $[-T,0]$. On the other hand, the graphs can never cross because at the first moment when $w_1=w$, we must have $\dot w_1<\dot w$, so a tiny bit before that we must have $w_1>w$, which contradicts the assumption that our crossing moment is the first one.
Observation 3. Suppose that $H$ is a canonical ascending regime and $w$ is the corresponding canonical solution with $\dot w\ge 0$. Suppose also that $c>0$ (it can be arbitrarily small) and $w_1$ satisfies $w_1^4+\dot w_1^2>(1+c)H$ with on $[-T,T]$. Then $w_1\ge w$ on $[0,T]$ if $T>T(c)$. Indeed, we have $w_1(0)\approx (1+c)^{1/4}H_0^{1/4}>w(0)$. Again, consider the first crossing moment if it exists. At that moment we must have $|\dot w_1|>\dot w$.
If $\dot w_1>0$, we get the contradiction in the same way as before. However, if $\dot w_1<0$, $w_1$ must continue to head down and its derivative can only grow in absolute value in that process until the direction gets reversed so this situation corresponds to a prohibited regime.
Observation 4. The time in the equation is reversible, so instead of investigating the descending regimes, we can investigate the ascending ones. The question we'll ask will be the following. Take any canonical regime $H$. Suppose we have a clever control $v$ such that $v(t)$ depends only on $H(s)$ with $s\ge t$ (there are no assumptions about how you designed it: solving the ODE, writing some explicit formula, asking an oracle, whatever). All we know is that our control doesn't create relative error more than $\delta$ for every regime anywhere and it is not allowed to see the past (remember that the time is now running backwards). What can we say about the value $v(t)$ for some fixed $t>0$?
The catch is that now we can glue any canonical regime on $[-T,t]$ which has correct values $H(t)$ and $\dot H(t)$ to the piece we have on $[t,T]$ and $v(t)$ is determined by that piece alone. Thus, $v(t)$ should make sense for any piece we glue.
Observation 5. Let us glue a canonical piece $H=w^4+\dot w^2$ with $\dot w\ge 0$ on $[-T,t]$ (note that we can choose $w$ instead of choosing $H$). Then the solution $(1+\delta)^{1/2}w$ will give us a canonical regime that is at least $(1+\delta)H$. Similarly, $(1-\delta)^{1/2}w$ will give us a canonical regime that is at most $(1-\delta)H$. Thus, our clever control must satisfy $(1-\delta)^{1/2}w(t)\le v(t)\le (1+\delta)^{1/2} w(t)$. Note that it should hold for every admissible $w$. Now you, probably, smell trouble: we have many different $w$ to try and, if they don't give the same value, the exact solution is impossible. Moreover, if those values may be noticeably different, $\delta$ cannot be too small.
Observation 6. We still have two conditions to satisfy at $t$, which, in terms of $w$, can be written as $w^4+\dot w^2=H$, $2w^3\dot w(2+\psi)=\dot H$ where $\psi=\ddot w/w^3$.
Note that we have 3 free parameters $w,\dot w,\ddot w$ and only 2 equations. Now it remains to find the quantitative estimates for the freedom we have. To this end, we will compare the scale invariant quantities $I=w^4/H$ for two regimes with the same $\dot H/H^{5/4}$. Since the controller I gave creates dismal results near the bottom of the cosine wave $2+\cos kt$ for $k>2$, it will be a good idea to look at the "quadratic departures" from the ground state and compute the maximal uncertainty along the way.
For technical reasons, it'll be convenient to look at $w(t)=1+at^2$, i.e., $H(t)=(1+at^2)^4+4a^2t^2$. The value $k=2$ corresponds to $a\approx 0.35$ here. This regime gives the invariant derivative quantity $\frac{8at(1+at^2)^3+a}{[(1+at^2)^4+4a^2t^2]^{5/4}}$. This is to be compared with the time/scale invariant regime $w(t)=b/t$, i.e., $H(t)=\frac{b^4+b^2}{t^4}$ with the invariant derivative quantity $\frac{4}{(b^4+b^2)^{1/4}}$. The ratio of the corresponding $I$-values is
$$
\frac{(1+at^2)^4(b^4+b^2)}{[(1+at^2)^4+4a^2t^2]b^4}
$$
where $b$ is determined from the equation
$$
b^4+b^2=\frac{[(1+at^2)^4+4a^2t^2]^{5}}{[8at(1+at^2)^3+a]^4}
$$
I will leave it to you to run a simple script maximizing over $t$ ($t=0.2$ already gives you something to be unhappy with for $a\in(0,1)$) and to recast the corresponding "inevitable uncertainty ratio" into the $\delta$-values. The table is like that:
$a=0.1, \delta=0.006$
$a=0.2, \delta=0.018$
$a=0.3, \delta=0.036$
$a=0.4, \delta=0.058$
$a=0.5, \delta=0.085$
$a=0.6, \delta=0.116$
$a=0.7, \delta=0.150$
$a=0.8, \delta=0.188$
$a=0.9, \delta=0.228$
As you can see, at the "breakdown point" $a=0.35$, the error of $4\%$ in the $H$-profile approximation is just inevitable and my controller is about that accurate. If $a$ gets to $0.6$, which corresponds to $k=\sqrt{8(a+a^2)}=2.8$ in the cosine model, you are going to be $10\%$ off somewhere no matter what you do. So, the version of the problem you are trying to solve is not just hard but unsolvable. You have to restrict the input to something reasonable and be happy with approximate control. However, you can try to investigate what features of the current control you dislike and design a better control in some respects. Just don't waste your time trying to achieve the impossible :).