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][1] 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][2] 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][3]. 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, \[PRL, 94, 176805 (2005)\]][4].

Yet Another Update: Apparently you have the ["Signum-Gordon" equation][5]. Thanks to [this answer][6].


  [1]: https://en.wikipedia.org/wiki/Quartic_interaction
  [2]: https://en.wikipedia.org/wiki/Sine-Gordon_equation
  [3]: https://en.wikipedia.org/wiki/Toda_lattice
  [4]: https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.94.176805
  [5]: https://www.actaphys.uj.edu.pl/index_n.php?I=R&V=38&N=10#3099
  [6]: https://mathoverflow.net/questions/22514/sign-gordon-equation/22590#22590