vibrating charged string
Let us, for concreteness, talk about the string with fixed endpoints, and one with a homogeneous, fixed charge density along it. When we derive the string equations, we essentially use the assumption $$k A \ll L$$ where $k$ is the number of antinodes ($k\geq 1$), $A$ is the amplitude of the oscillation, and $L$ the string length. Then, if we take the leading order term $\sim kA/L$ in our equations and neglect anything $\mathcal{O}(k^2 A^2/L^2)$, we get the linear string equation as you know it. Thus, we should only consider the effects of lowest order in $A/L$ in our computation.
Now let us assume that the charge density is so small that the string does not yet feel any back-reaction. Then we make a multipole expansion of the electromagnetic radiation leaving the system and we obtain that the average power radiated by the $2^n$ multipole in all directions is $$P_{2^n} \propto \frac{1}{\lambda^2} \frac{|M_{2^n}|^2}{\lambda^{2n}}$$ where $\lambda$ is the wave-length of the electromagnetic radiation, and $|M_{2^n}|$ is the amplitude of the oscillation of the $2^n$ pole (assuming it undergoes a harmonic oscillation, also note that there are "longitudinal" parts of the multipoles in the string which do not oscillate). On the other hand, the time-varying $2^n$-pole is proportional to the total charge of the object $Q$ and some typical length to the n-th power, in our case you can compute that the time-variable part will be always proportional to $A^n$. (If you are unsure, take a look at the $n=1$ dipole, and $n=2$ quadrupole radiation computation on wiki.)
The frequency of the emitted radiation $f_{EM}$ will be some integer multiple of the frequency of the vibration of the string, which is $$f_s = \frac{k v_s}{2L}$$ where $v_s$ is the wave-speed in the string. Since $\lambda=c/f_{EM}$, after dropping some integer factors we finally get $$P_{2^n} \propto \frac{Q^2}{\lambda^2} \left(\frac{A}{L} \frac{v_s}{c}\right)^{2n}$$ That is, the various multipole contributions to the radiation of the string drop of very quickly with exactly the expansion parameter $A/L$ whose higher powers we neglect in the expansion. On the other hand, you can verify that the energy of the string is $\propto (A/L)^2$ itself, so in a consistent $(A/L)$-power expansion we necessarily get that the dipole radiation is stealing energy already in the linear mode of the oscillations of the string. Then again, unless the back-reaction becomes strong, it is safe to assume that the quadrupole and higher-order leakage of energy will be negligible even when we assume some back-reaction on the string (the dimensional analysis makes this more or less inevitable).
Now, this doesn't quite help us to solve the nonlinear back-reaction equations as you propose in your question, but at least it allows us to build a leading-order model for the behaviour of the string. This is because we know from the dipole radiation formula the loss of energy averaged over one period must be (now adding all the factors) $$-\langle \frac{d E_s}{d t} \rangle = P_2 = \frac{\mu_0 }{12 \pi c} \omega^4 |M_2|^2 $$ Now let us assume that the change of the amplitude of a given mode $A_k$ is so slow that we can use this average to compute the evolution of it $A_k$ by simply integrating it. We use the formula for the energy of the mode $E_{ks} = M A_k^2 \omega_k^2/2$ where $M$ is the string mass, $\omega_k = \pi v_s k/L$. A quick computation also yields $M_{2,k} = Q/k$ for k odd, and $M_{2,k} = 0$ for k even, where $Q$ is the total charge of the string. When the dust settles, you obtain the average evolution of the amplitude of the $k-th$ mode is given by $$ \dot{A}_k = - \zeta A,\;k\;odd$$ where $\zeta$ is a constant damping rate I am sure you will be able to retrieve from the equations above. (The validity of the "slow-change" approximation is $\zeta \ll \omega_k$.) The important message is that the amplitude will evolve as $$A_k(t) \propto e^{- \zeta t},\;k\;odd$$ and, quite surprisingly, the even modes will not change, at least on the time-scale where the linear string approximation will be valid.
So this was a short treatment which shows you that the dominant reaction of the string will be to radiate away its dipole oscillation. Of course, once non-linearities start to show up, internal physics of the string transferring energies from mode to mode as well as the higher-order multipole and non-trivial back-reaction effects start to show. However, notice that "wave-like" back-reaction does not show in the linear-string mode since the string elements move at speeds $\sim v_s A/L \ll c$. So in the linear mode we could only hope for a slightly better description of the evolution of $A_k$ similar to the damped harmonic oscillator.
Another problem is that the back-reaction of small singular objects such as infinitesimal strings is generally ill-behaved, and it is better to state what is the model really describing and compute from there. A similar problem would be with the $\mathcal{O}(A^2/L^2)$ non-linearities, they will once again depend on the matter model. So, this is really as far as you can go with a self-consistent description of the problem.
Last note to the Abraham-Lorentz force - of course, that is not the only contribution, the string element would receive contributions from the electromagnetic field of its neighbours. I would guess that these contributions are actually dominant and that is the reason why the forces on some modes add up to damping while not for others.
I'm going to mostly focus on fields generated by the string. If you assume the string will have the traditional fixed end standing wave solutions, then we can describe the motion of the string as:
$$z(x,t)=A\sin{(\frac{\pi n}{L}x)}\cos{(\omega t)}$$
Each infinitesimal slice of the string looks like a charge:
$$ dq = \rho_L dl$$
Where $\rho_L$ is the linear charge density of the string. Furthermore, the infinitesimal dipole moment is then:
$$ dp = 2z(x,t)dq =2A\rho_L\sin{(\frac{\pi n}{L}x)}\cos{(\omega t)}dx\hat{z}$$
I think it's demonstrative (and certainly much easier) to see what would happen for the case where we are sufficiently far from the string (R>>L)
$$ \vec{p}=\int_0^Ldp=\frac{2AL\rho_L}{\pi n}\cos{(\omega t)}\hat{z} $$ for n = 1,3,5... and 0 for n = 2,4,6... This makes sense because if there are an even amount of modes they would effectively cancel each other out. So to find the fields, just take the electric and magnetic fields generated from a Hertzian dipole and replace the dipole moment with:
$$ \vec{p}_0 = \frac{2AL\rho_L}{\pi n}\hat{z} $$
Solving this without the R>>L assumption would require you to go back to solving the magnetic vector potential:
$$\vec{A}=-\frac{i\mu_0 \omega}{4\pi}\int_0^L\frac{e^{ikr}}{r}d\vec{p}$$
If you're just interested in the far-fields, I bet you could actually simplify it by solving it with a continuous array factor, the same way we solve for dipole antennas with non-uniform current.