As the other poster said, the thing for which you use $E$ in your expression is not the eigenvalue of the Hamiltonian, but rather the eigenvalue of the kinetic part of the Hamiltonian.
The Hamiltonian is:
$$ \hat H = \frac{\hat p^2}{2m} \otimes \mathbb{1} + \mathbb{1} \otimes h_z\sigma_z$$
and the eigenvalues are:
$$E_\pm(p) = \frac{p^2}{2m}\pm h_z$$
However, this doesn't really answer your question of are spin degrees of freedom physical in an absolute sense or merely relative; the answer is that the spin degrees of freedom are physical in an absolute sense, but several initial results suggest that they're not.
Firstly, measurement: the results of a measurement of spin could be anything you like, so there is no absolute meaning to spin in that context. Consider the following operator:
$$ \hat A = a_\uparrow\left|\uparrow \right\rangle\left\langle \uparrow\right| + a_\downarrow\left|\downarrow \right\rangle\left\langle \downarrow\right| = \left( \begin{array}{ccc}
a_\uparrow & 0 \\
0 & a_\downarrow \end{array} \right) $$
This is hermitian and has eigenvalues $a_\uparrow$ and $a_\downarrow$, which can be any real numbers as long as they aren't the same, so the results of a measurement of spin can be anything you like.
This is actually a generic properties of measurement in quantum mechanics, that the results can be relabeled in any way you like, so it doesn't really tell us anything about the physical significance of spin.
Next: time evolution of the spin state in a magnetic field. Let's consider if the values of the spin operator had some constant offset, ie:
$$ \hat S = S_0 + \frac{\hbar}{2}\sigma_z $$
Now the Zeeman terms in the Hamiltonian:
$$ \hat H _Z = a \hat S = a S_0 + a \frac{\hbar}{2}\sigma_z $$
where $a$ is some constant. Now consider the time evolution of a general state:
$$ \hat U(t)\left(\alpha \left|\uparrow \right\rangle + \beta \left|\downarrow \right\rangle \right) = \alpha \exp\left(\frac{aS_0t}{\hbar}+\frac{at}{2}\right)\left|\uparrow \right\rangle + \beta \exp\left(\frac{aS_0t}{\hbar}-\frac{at}{2}\right)\left|\downarrow \right\rangle $$
$$ = \exp\left(\frac{aS_0t}{\hbar}\right)\left(\alpha \exp\left(+\frac{at}{2}\right)\left|\uparrow \right\rangle + \beta \exp\left(-\frac{at}{2}\right)\left|\downarrow \right\rangle\right) $$
So the offset in the spin operator only contributed to a total phase offset, which makes no difference to the probabilities of spin measurements. This supports what you suggest---that only t difference in spin is physically relevant.
The absolute relevance of spin degrees of freedom, however, comes when you consider the intrinsic magnetic moment of particles with spin. The magnetic moment operator for a spin half particle is given by:
$$ \hat\mu_z = g\hat S_z=g\frac{\hbar}{2}\hat\sigma_z$$
where $g$ is the gyromagnetic ratio of that particle. The physical significance of this can be seen, for instance, in Stern-Gerlach experiments.
In these experiments you have a state where, in general, the position degrees of freedom are entangled with the spin degrees of freedom, ie:
$$ \left| \Psi \right\rangle = \alpha|\psi_\uparrow\rangle \otimes \left| \uparrow \right\rangle + \beta| \psi_\downarrow \rangle \otimes \left| \downarrow \right\rangle $$
I can't easily go over the derivation of the effect here, but the short story is that the evolution of the spatial state depends, via the entanglement, on the direction of the spin state to which the spatial state is entangled. Then the final position of the particle on the screen depends on the spin, hence the experiment constitutes a measurement of spin.
The final position of a wavepacket does depend on the absolute value of spin, giving absolute physical significance to the spin degrees of freedom.
