Maxwell Equations don't give unique Electric Field?
The problem is that you calculated the curl wrong; you missed a delta function arising from the discontinuity. We can write $E_z$ as
$$ E_z = \ln(Cr) \Theta(R-r)$$
and, using that $\Theta'(x) = \delta(x)$, we get that
$$\nabla \times \mathbf{E} = -\frac{1}{r} \Theta(R-r) \hat{\phi} - \ln(CR)\delta(r-R)\hat{\phi}$$
so when giving $\nabla \cdot \mathbf{E}$ and $\nabla \times \mathbf{E}$ you do need to specify $C$: it's not a free constant.
This can be illustrated with a much simpler example. Suppose you demand $\nabla \cdot \mathbf{E} = 0$, $\nabla \times \mathbf{E} = 0$ and $\mathbf{E} \to 0$ at infinity. Then if you forget about the delta function, any field which is constant in a bounded region of space and zero elsewhere solves these equations.
The moral is that to get uniqueness out of Maxwell's equations you either need the fields to be differentiable (or maybe $C^1$), which your example is not, or you need to use distributions like I did.
Regarding your last point: there is no reason a given magnetic field should produce a unique electric field, because knowing $\mathbf{B}$ determines $\nabla \times \mathbf{E}$ and $\partial \mathbf{E} / \partial t$ but not $\nabla \cdot \mathbf{E}$ or the boundary conditions. You could add any static field with zero curl and still get a valid solution.
Let's see what kinds of sources could produce these fields. I rewrote this part because it was kind of a mess.
First observe that since $\nabla \cdot \mathbf{E} = 0$, there are no charges. We could have missed charges at $r=0$ because $\mathbf{E}$ is no defined there, but integrating along a cylindrical Gaussian surface reveals that there are no charges there either.
For later we'll need the following:
$$-\nabla^2 \mathbf{E} = \left[ \frac{\delta(r-R)}{r}(2+\ln(CR)) + \delta'(r-R)\ln(Cr)\right] \hat{z} \equiv \epsilon(r)\hat{z}$$
Using Faraday's law with zero magnetic field at $t=0$ we get that
$$\mathbf{B} = -(\nabla\times\mathbf{E})t = \left[\frac{1}{r}\Theta(R-r) + \ln(CR)\delta(r-R)\right] t \hat{\phi}$$
Now we can get the currents from Ampère's law. The differential version fails to reveal a wire of current at $r=0$ because $\mathbf{B}$ is undefined, so we need to use the integral version for that. The full result is
$$\mathbf{J} = \left[\frac{\delta(r)}{r}\hat{z}-\nabla^2 \mathbf{E}\right]t = \left[\frac{\delta(r)}{r}+\epsilon(r)\right]t\hat{z}$$
There are infinities everywhere, but as we know these aren't really a problem if we interpret them as approximations of smooth fields and currents. We have a wire at $r=0$ that carries a current that goes linearly with $t$; this is prefectly reasonable. The behavior at $r=R$ is a little weirder. There is a delta-function current sheet and a delta-function-derivative current sheet, which we might interpret as two currents going opposite ways very close together, not unlike a dipole sheet in electrostatics.
Conclusion: as long as it is understood that infinite current densities are really approximations of very thin current sheets and that the solutions hold for a finite time interval (as they always do), this solution seems physical enough. At least it's not completely unrealistic. You might have some trouble with all the overlapping current sheets if you try to make it in the lab, though.
First of all, the tangential component of ${\bf E}$ (i.e. the component parallel to the interface) is always continuous across interfaces, so $E_z$ must be continuous at $r = R$, which fixes $C = 1/R$. There is no physical configuration of sources that could produce any other value for $C$.
${\bf E}$ does not depend on time, so ${\bf \nabla \times E} = -\partial {\bf B}/ \partial t$ gives $$ {\bf B} = -({\bf \nabla \times E})\, t = -\frac{1}{r} \Theta(R - r)\, t\, \hat{\phi} $$ (choosing for simplicity to set ${\bf B}(t = 0) = {\bf 0}$), so the magnetic field is azimuthal and increases steadily with time inside the cylinder of radius $R$.
Taking the curl gives that for $r > 0$, $$ {\bf \nabla} \times {\bf B} = {\bf J} = \frac{1}{r} \frac{d (r\, B_\phi)}{d r} \hat{z} = \frac{\delta(r-R)}{r}\, t\, \hat{z}.$$
We have to consider the axis $r = 0$ separately, because the fields diverge there. Noting that $\oint {\bf B} \cdot d{\bf l} = -2 \pi t$ for a circle around the $z$-axis inside radius $R$, the integral form of Ampere's law gives that there is a current ${\bf J} = -t\, \hat{z}$ at the axis.
The physical interpretation is that there is a line of electric current running down the $z$-axis and a cylindrical sheet of current with radius $R$ running up, both of them steadily increasing in time, with no net current parallel to $\hat{z}$. It's kind of like a solenoid with a square cross-section wound into a donut, then the hole in the middle shrunk down to a line, then the two flat faces taken very far away - and just like a solenoid, a steadily increasing current induces a steadily increasing ${\bf B}$ field but a constant ${\bf E}$ field.
Maxwell's equations only give a unique electric field subject to a set of boundary conditions and an initial condition for the field.