The Lagrangian (and the action as a whole)
$$ L = \frac{1}{2}m \dot{q}^2 - \ln t$$
is not invariant under the transformation given by
$$T = t, \qquad Q = 0.$$
The rescaling of $t$ by factor $1+\epsilon$ also modifies the time derivatives:
$ \delta \dot{q} = -\epsilon \dot{q}$ (and the measure of integration $dt$), so the suggested quantity is not conserved.
Where is my error?
So, the error is in choosing the wrong Lagrangian / transformation.
Noether's theorem works just fine for explicitly time dependent Lagrangians.
Here is another example of Lagrangian with explicit time-dependence:
$$
L = \frac{m \dot{q}^2}{2} e^{\alpha t}.
$$
Such type of Lagrangian could arise for example, if we try to obtain the equations of motions for dissipative system.
The Euler-Lagrange equation for this system after omitting common factor reads as
$$
\ddot{q} = -\alpha \dot{q}.
$$
This Lagrangian is invariant under the infinitesimal transformation:
$$
t\to t' = t + \epsilon, \qquad q \to q' = q -\epsilon\frac{\alpha q}{2}.
$$
Substituting these $T$ and $Q$ in the Noether's theorem we obtain the quantity
$$
A=\frac{m }{2} e^{\alpha t}\cdot (\dot{q}^2 + \alpha \dot{q} q).
$$
Its time derivative is
$$
\dot{A} = \frac{m }{2} e^{\alpha t}\cdot (\alpha \dot{q}^2 + \alpha^2 \dot{q} q +2 \ddot{q}\dot{q} + \alpha \ddot{q} q + \alpha \dot{q}^2) ,
$$
If we use E-L equation to eliminate $\ddot{q}$ we obtain $\dot{A}=0$, so the quantity is conserved as it should be.
A translation by $x^\nu \to x^\nu - \epsilon^\nu$ corresponds to an infinitesimal transformation of the fields, by
$$\phi \to \phi + \epsilon^\nu \partial_\nu \phi$$
as we are performing an active rather than passive transformation. The Lagrangian transforms as,
$$\mathcal{L}\to \mathcal{L}+\epsilon^\nu \partial_\nu \mathcal{L}$$
by substituting $\phi$ into the Lagrangian. Notice the change is up to a total derivative, and hence Noether's theorem is applicable to the symmetry. The conserved current density is given by,
$$j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}X(\phi)-F^\mu(\phi)$$
where $X=\delta\phi$ and $F^\mu$ is such that $\partial_\mu F^\mu=\delta \mathcal{L}$ infinitesimally. For our case, we obtain the symmetric stress-energy tensor (analogous to that of general relativity),
$$T^\mu_\nu=\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \partial_\nu \phi - \delta^\mu_\nu \mathcal{L}$$
where the Kronecker delta is raised with the Minkowski metric. The current satisfies, $\partial_\mu T^{\mu}_\nu = 0$, and the corresponding Noether charge,
$$E=\int \mathrm{d}^3 x \, T^{00}$$
is the total energy of the system, whereas,
$$P^i = \int \mathrm{d}^3 x \, T^{0i}$$
is the $i$th component of the total momentum of the field, where $i=(x,y,z)$ only. A caveat: the stress-energy tensor derived by Noether's theorem is not always symmetric, and may require the addition of a term which satisfies the continuity equation, and ensures symmetry in the indices.
Alternate Method
Recall to obtain the Einstein field equations in general relativity, we may vary the Einstein-Hilbert action,
$$S\sim \int \mathrm{d}^4 x \, \sqrt{-g} \, \left( R + \mathcal{L}\right)$$
Similarly, in quantum field theory, we may promote our Minkowski metric to a generic metric tensor, thereby replacing the kinetic term of the Lagrangian with covariant derivatives. Up to some constants, the stress-energy tensor is given by
$$T^{\mu\nu} \sim \frac{1}{\sqrt{-g}} \frac{\partial (\sqrt{-g}\mathcal{L})}{\partial g^{\mu\nu}}$$
evaluated at $g_{\mu\nu}=\eta_{\mu\nu}$, which is precisely the definition we implement when obtaining the Einstein field equations for general relativity.
Best Answer
Here I would like to mention the notion of quasi-symmetry. In general, if the Lagrangian (resp. Lagrangian density resp. action) is only invariant up to a total time derivative (resp. space-time divergence resp. boundary term) when performing a certain off-shell$^1$ variation, one speaks of a quasi-symmetry, see e.g. Ref. 1.
Noether's first Theorem does also hold for quasi-symmetries. For examples of non-trivial conservation laws associated with quasi-symmetries, see examples 1, 2 & 3 in the Wikipedia article for Noether's theorem.
References:
--
$^1$ Here the word off-shell means that the Euler-Lagrange (EL) eqs. of motion are not assumed to hold under the specific variation. If we assume the EL eqs. of motion to hold, any variation of the Lagrangian is trivially a total derivative.