For continuous global symmetries, Noether theorem gives you a locally conserved charge density (and an associated current), whose integral over all of space is conserved (i.e. time independent).
For global discrete symmetries, you have to distinguish between the cases where the conserved charge is continuous or discrete. For infinite symmetries like lattice translations the conserved quantity is continuous, albeit a periodic one. So in such case momentum is conserved modulo vectors in the reciprocal lattice. The conservation is local just as in the case of continuous symmetries.
In the case of finite group of symmetries the conserved quantity is itself discrete. You then don't have local conservation laws because the conserved quantity cannot vary continuously in space. Nevertheless, for such symmetries you still have a conserved charge which gives constraints (selection rules) on allowed processes. For example, for parity invariant theories you can give each state of a particle a "parity charge" which is simply a sign, and the total charge has to be conserved for any process, otherwise the amplitude for it is zero.
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.
Best Answer
Nope, this operation is not a symmetry in the physics sense. A symmetry transformation is a transformation that changes or mixes the values of the basic degrees of freedom such as positions and momenta $x(t)$, $p(t)$ in mechanics or the values of fields such as $\vec E(\vec x,t)$ and $\vec B(\vec x,t)$ in electromagnetism.
The Hamiltonian is not an independent variable or a basic degree of freedom; it is a function of them. You're not changing the values of any quantities that evolve with time; instead, you're changing some formulae for derived and in principle unnecessary auxiliary objects in the theory (in this case the Hamiltonian), claiming that other formulae are preserved. This is not a symmetry so there is no conservation law associated with this operation. You're not "rotating" real objects which is what symmetry transformations should do: you're just redefining auxiliary, derived objects on the paper.
Incidentally, the operation you mention fails to be "harmless" in general relativity because the energy is a source of gravitational field - curvature of space - so if you move it by a constant, you do change physics.