This answer really has to do with the physics of it: are you sure about the area energy term?
Let me simplify a little bit: Consider a case where $z_1(x)=0$ and $z_2(x)=w(x)$. According to your formula I should get a term proportional to $w'(x)^2$ for the kinetic term which is typical and no one will object to you for that. You potential energy would then be proportional to $\int_0 ^L |w(x)| dx$. That does raise an alarm: nonlinear problem.
Unless you really have nonlinear physics going on (which is likely, see below) you should have had $w(x)^2$. That would bring back the $L_2$ norm and everything will be very simple. Just transform $z_2(x)=z_1(x)+w(x)$, get rid of $z_2$ and you will get two uncoupled (continuum of) modes, one (corresponding to $z_1$) is a free particle and the other a Harmonic oscillator. Your Boltzmann statistic will determine how these modes are filled up as you know.
If you think of your rubber band as a collection of springs and masses, $w(x)^2$ is the actual term but collections of springs and masses hardly exist outside textbooks and problem sets. As a physicist you know: molecules interact. The actual expansion (if you are still interested in writing an effective field theory) will involve (interacting) terms of the type $w(x)^2+O[w(x)^4]$. Quantitative results in this model might involve renormalization. See the wikipedia page on quartic interaction that even describes how you should quantize it (Fourier transform). My field theory is rusty so you might already know more than I do.
What if $|w(x)|$ is what you have: If $|w(x)|$ is not too small or too large, just approximate it with $[w(x)]^2$ to linearize the problem. People might point out that this is not a mathematically good approximation but it will physically make sense: very small stretchings are not physically possible and very large stretchings would be ruled out by the Boltzmann statistics as they would correspond to exponentially rare high energy modes. So I would just introduce a factor so that $w(x)$ and $w(x)^2$ coincide where $\kappa w(x)^2 L\approx kT$.
There are a famous nonlinear equations that come cheerfully close to your problem but miss it. Example: Sine-Gordon Equation where instead of $|w(x)|$ you would have $1-\cos [w(x)]$. Actually more up to the point would be the Sinh-Gordon... which reminds me of the Toda field theory which describes a Toda lattice. A Toda lattice is a nonlinear set of coupled equations that describe a set of nonlinearly coupled particles. If you are working in the liquid state, I doubt that they would be relevant. $w(x)^2$ should be good enough but the dissipative terms will be more troublesome.
Edit: Couldn't resist the pun. Seems that the relevant equation is a "SIGN-Gordon equation": $$\varphi_{tt}- \varphi_{xx} + \text{sgn}(\varphi) = 0.$$ Not sure if it is really simple, messy, or plain difficult to solve in your case. An option is to try solve it like a wave equation with a sign changing external force term and then parametrize the solutions based on their energy and apply the Boltzmann statistics. Forgo the Hamiltonian altogether. I think it will be messy.
Another Edit: as the commenters mentioned before, the solution to the absolute value potential Schrodinger eqn. involves Airy functions. So you could in principle use the Airy function eigenstates to find the time-independent modes and fix their energy. This could quantize your problem. The solution in terms of the Airy functions can be found for example here, [PRL, 94, 176805 (2005)].