Can Noether's theorem be understood intuitively?

It's intuitively clear that the energy most accurately describes how much the state of the system is changing with time. So if the laws of physics don't depend on time, then the amount how much the state of the system changes with time has to be conserved because it's still changing in the same way.

In the same way, and perhaps even more intuitively, if the laws don't depend on position, you may hit the objects, and hit them a little bit more, and so on. The momentum measures how much the objects depend on space, so if the laws themselves don't depend on the position on space, the momentum has to be conserved.

The angular momentum with respect to an axis is determining how much the state changes if you rotate it around the axis - how much it depends on the angle (therefore "angular" in the name). So the symmetry is linked to the conservation law once again.

If your intuition doesn't find the comments intuitive enough, maybe you should train your intuition because your current intuition apparently misses the most important properties of time, space, angles, energy, momentum, and angular momentum. ;-)

The intuitive argument for Noether's theorem, which is also the best completely precise argument for Noether's theorem, appears in Feynman's popular book "The Character of Physical Law". I will reproduce the argument which relies on the following diagram:

In this diagram the two parallel squiggles with a line connecting them at the top and at the bottom represent a particle path and a displaced particle path.

The action is stationary on the particle path, so the square squiggle which translates over, goes up parallel, and comes back has the same action as the original path. The original path, however, has the same action as just the squiggle part of the other path, therefore the two horizontal lines at top and bottom have equal action.

You can use this argument to find the exact form of the Noether current by replacing Feynmans horizontal lines with quick kicks by the momentum over a time $\epsilon$. His argument is an honest to goodness proof, it is by far the best proof, and it is the only case in all the history of publishing where a result is best presented in a popular book.

If you make the kicks continuous in time, so that they come here and there, you can still see that the kicks integrate by parts. This argument appears in the introduction to one of Hawking's 1970s papers, and is essentially equivalent to Feynman's "Character of Physical Law" argument, except it appears more than ten years later.

Well, I don't know about any intuitive explanation besides intuition gained by understanding the underlying math (mainly differential geometry, Hamiltonian mechanics and group theory). So with the risk of not giving you quite what you want, I'll try to approach the problem mathematically.

If you know Hamiltonian mechanics then the statement of the theorem is exceedingly simple. Assume we have a Hamiltonian $H$. To this there is associated a unique Hamiltonian flow (i.e. a one-parameter family of symplectomorphisms -- which is just a fancy name for diffeomorphisms preserving the symplectic structure) $\Phi_H(t)$ on the manifold. From the point of view of Lie theory, the flow is a group action and there exists its generator (which is a vector field) $V_H$ (this can also be obtained from $\omega(\cdot, V_H) = dH$ with $\omega$ being the symplectic form). Now, the completely same stuff can be written for some other function $A$, with generator $V_A$ and flow $\Phi_A(s)$. Think of this $A$ as some conserved quantity and of $\Phi_A(s)$ as a continuous family of symmetries.

Now, starting from Hamiltonian equation ${{\rm d} A \over {\rm d} t} = \left\{A,H\right\}$ we see that if $A$ Poisson-commutes with $H$ it is conserved. Now, this is not the end of the story. From the second paragraph it should be clear that $A$ and $H$ don't differ that much. Actually, what if we swapped them? Then we'd get ${{\rm d} H \over {\rm d} s} = \left\{H,A\right\}$. So we see that $A$ is constant along Hamiltonian flow (i.e. conserved) if and only if $H$ is constant along the symmetry flow (i.e. the physical laws are symmetric).

So much for why the stuff works. Now, how do we get from symmetries to conserved quantities? This actually isn't hard at all but requires some knowledge of differential geometry. Let's start with most simple example.

Translation

This is a symmetry such that $x \to x^\prime = x + a$. You can imagine that we move our coordinates along the $x$ direction. With $a$ being a parameter, this is a symmetry flow. If we differentiate with respect to this parameter, we'll get a vector field. Here it'll be $\partial_x$ (i.e. constant vector field aiming in the direction $x$). Now, what function on the symplectic manifold does it correspond to? Easy, it must be $p$ because by differentiating this we'll get a constant 1-form field $dp$ and then we have to use $\omega$ to get a vector field $\partial_x$.

Other way to see that it must be $p$: suppose you have a wave $\exp(ipx)$. Then $\partial_x \exp(ipx) = ip \exp(ipx)$ so momentum and partial derivatives are morally the same thing. Here we're of course exploiting the similarity between Fourier transform (which connects $x$ and $p$ images) and symplectic structure (which combines $x$ and $p$).

Rotation

Now onto something a bit harder. Suppose we have a flow $$\pmatrix{x \cr y} \to \pmatrix{x' \cr y'}= \pmatrix{\cos(\phi) & \sin(\phi) \cr - \sin(\phi) & \cos(\phi)} \pmatrix {x \cr y} $$ This is of course a rotational flow. Here we'll get a field $y {\rm d}x - x {\rm d} y$ and the conserved quantity of the form $y p_x - x p_y$ which can in three dimensions be thought of as a third component of angular momentum $L_z$.

Note that the above was done mainly for illustrative purposes as we could have worked in polar coordinates and then it would be actually the same problem as the first one because we'd get the field $\partial_{\phi}$ and conserved quantity $p_{\phi}$ (which is angular momentum).