Noether's theorem is overzealously applied--- it only applies to theories with a Lagrangian formulation, or to quantum mechanics. This is true of fundamental systems, but for non-fundamental systems, you can have classical equations which are symmetrical the symmetry does not imply a conservation law.
The symmetry does not come with no consequence, however, it comes with the consequence of symmetrical solutions! If the equations are symmetrical under a transformation, the solutions must come in families that turn into each other under the symmetry. For classical systems, this is not a particularly profound consequence. So I will consider systems where this is the only consequence.
For a stupid example, consider Newton's laws for an object free-falling in a gravitational field. The acceleration is uniform in the z direction, but z momentum is not conserved. The reason is that the Lagrangian is not invariant in the z direction. But you wouldn't know it from looking at the equations of motion.
For a less stupid example, consider Newton's laws for a particle with constant force and a $v^3$ friction law, say:
$$ {dv \over dt} = a + v^3 $$
And there is a symmetry in time translation and for translations in x. But aside from telling you the trivial fact that solutions are translatable in x and in t, it doesn't tell you anything more.
These problems are sort of silly, so I'll give the granddaddy of all examples--- the incompressible Navier stokes equations, with hyperviscosity (so that what I am saying is definitely true). Here you have a time-dependent diffeomorphism $X(x)$ from n-dimensional space whose general point is called x, to itself.
The time derivative of X is the velocity field v, and v obeys the equation
$$ {\partial_t} v = v\cdot \nabla v + \nabla P + \nu \nabla^2 v + \epsilon \nabla^4 v $$
Where the $\epsilon$ term is introduced to make sure that the equation has a unique and smooth initial value problem, so that the X diffeomorhism makes sense. This equation is translationally invariant to t translations, and completely diffeomorphism invariant--- composing X with a diffeomorphism takes a solution to a solution, but it doesn't have a conserved energy (although formally the limit $\nu=\epsilon=0$ does), nor does it have any conserved quantity corresponding to the diffeomorphism invariance. The diff invariance is a gauge redundancy in the X description.
These symmetries still have the trivial consequence of symmetrical solutions--- so you can translate any solution in time, and do a diffeomorphism on the initial positions. It just doesn't mean anything to studying the equation.