In QFT, do the fields evolve with determinism, in principle? In quantum mechanics, the outcomes of a certain measurement might not be deterministic. However, the wavefunction evolves with determinism according to Schrodinger's equation.
Is QFT analogous in that aspect? Is the universe state represented by quantum field configurations which evolve with determinism according to some law of physics?
I feel the question is not very well coined, I can't help: I'm a mere newbie to this field.
 A: Some remarks:
1) Wavefunctions are not classical fields, and are not field operators.
2) Quantization transform classical fields in operators (so equations of movement, or equations of conservation of energy, are equations between operators).
3) Quantum Mechanics may be seen as Quantum Field Theory with zero spatial dimension.
4) Following the above remark, it is better to see the position of the particle, as a "field" (of time). So, you may, in QM, make the substitution $x \to \phi$, in all your textbooks formulae, including operators :($ \hat \phi$ instead of $\hat x$), bras, kets, and parameters of the wave function ($\psi(\phi, t)$ instead of $\psi(x, t)$). 
The usual QM Schrodinger equation is then : 
$- \dfrac{\hbar ^2}{2m} \dfrac{\partial^2 \psi(\phi, t)}{\partial \phi^2} + V(\phi)\psi(\phi, t)= i \hbar \dfrac{\partial \psi(\phi, t)}{\partial t}$
5) The generalization to QFT is to consider a wavefunction $\psi (\{\phi_k\},t)$.
To see that,  let's take the very simple case of  a free massless real scalar (bosonic) classical field $\phi(\vec x,t)$, with equation $\square \phi(\vec x,t)=0$, applying a spatial Fourier transform gives : $  \dfrac{\partial^2 \phi(\vec k, t)}{\partial t^2} + \vec k^2 \,\phi(\vec k, t)=0 $. This means that the collection $\phi_k(t) = \phi(\vec k, t)$ is a set of independent harmonic oscillators. So, we know how to quantize a harmonic oscillator, then, for each $\vec k$ mode, there is a wave function $\psi_k (\phi_k,t)$ verifying a Schrodinger-like equation: 
$ - \frac12 \dfrac{\partial^2 \psi_k(\phi_k, t)}{\partial \phi_k^2} + \frac12 \vec k^2 \,\phi_k^2\,\psi(\phi_k, t)= i  \dfrac{\partial \psi_k(\phi_k, t)}{\partial t}$
One may build a global wave function $\psi (\{\phi_k\},t) = \prod\limits_k \psi_k(\phi_k, t)$, with a Schrodinger-like equation: 
$ \sum\limits_k (- \frac12 \dfrac{\partial^2 }{\partial \phi_k^2} + \frac12 \vec k^2\,\phi_k^2\,)\psi (\{\phi_k\},t)= i  \dfrac{\partial \psi (\{\phi_k\},t)}{\partial t}$
6) However, speaking of fields, they are, in QFT, operators, noted by $\hat \phi_k(t)$,  these operators apply on states, and their equation is simply : 
$  \dfrac{\partial^2 \hat \phi_k(t)}{\partial t^2} + \vec k^2 \,\hat \phi_k( t)=0 $
As noted in $2)$, it is the same equation as classical fields, but for operators.
