The situation is a bit difficult if you want to take the strong interaction effects and so on into account, because it's a question what you permit to be "geometrical". The fundamental forces involving charges are decribed by qunatum field theories, gauge theories of the the Yang–Mills type and as far as this whole connection-business goes, it is actually a pretty geometric setting. So from a mathematical side, there are several similarities to general relativity. These theories are not concerned with a curved base manifold and the tangent space in the exact same sense as Riemann-geometry, but there is also a notiond of curvature. I don't know if these will help you but here are two more links regarding a united mathematical perspective.
For starter, lets keep this pre quantum. The obvious thing that comes to mind here is Kaluza Klein theory. However, as you'll read in the very first paragraph, this is a theory in five dimensions. If you don't mind this, then I'd push you further in a stringly direction, although this also differs pretty much from conventional general relativity. It's not unlikely that some people here might write an answer from that perspective.
The incorporation of a force like electromagnetism in a classical four dimensional Pseudo-Riemann geometry type theory like general relativity turns out to be problematic. Without going into computations involving features like spacetime metrics, Christoffel symbols and the possibility to choose a local inertial frame, here is an argument involving the related equivalence principle. I quote the weak, Einsteinian and strong version from wikipedia here and highlight some words:
Weak: "All test particles at the alike spacetime point in a given gravitational field will undergo the same acceleration, independent of their properties, including their rest mass."
Einsteinian: "The outcome of any local non-gravitational experiment in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime."
Strong: "The gravitational motion of a small test body depends only on its initial position in spacetime and velocity, and not on its constitution and the outcome of any local experiment (gravitational or not) in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime."
Let's consider a situation in general relativity where you have a big electrically negative charged object $X$ and the corresponding spacetime curvature is governed by the Reissner–Nordström metric. You are sitting in your cosy laboratory, falling through space and you have optionally two to four particles $A,B,C$ and $D$ of charge $Q_A=0, Q_B=-1, Q_C=+1$ and $Q_D=-9001$. Lets say at the beginning of your experiments they are never moving relative to you. You and the uncharged particle will fall freely (along a geodesic produces by the Reissner–Nordström metric) towards $X$, while the other particles are never free falling, interact electromagnetically with $X$ and therefore move in different ways, depending on their charge. They also might interact with each other in various ways. There is only one particle mass, but netral, positive and negative electric charges so that different species are affected differently by the surroundings.
Given the fact that the physical observations so far agree with the results of general relativity: If you incorporate elecromagnetism in your four dimensional curved spacetime of your geometrical theory, how would you realize the equivalence principle?
The notion of "conservative forces" is not in any way fundamental. What's fundamental is that we're able to assign a number to the state of a system, and that number is conserved.
The relatively uninteresting notion of a "conservative force" can be applied only to a force that can be expressed as a vector field that depends only on position. That means it's meaningful for Newtonian gravity and for electrostatics, but not for any other force that could be considered fundamental. Re the nuclear forces, see Do strong and weak interactions have classical force fields as their limits? .
General relativity has local conservation of energy-momentum, which is expressed by the fact that the stress-energy tensor has a zero divergence. A mass-energy scalar or energy-momentum vector isn't something that can be defined globally in GR for an arbitrary spacetime.
WP says (Gugg, where was the link from?):
However, general relativity is non-conservative, as seen in the anomalous precession of Mercury's orbit. However, general relativity can be shown to conserve a stress-energy-momentum pseudotensor.
The first sentence is wrong, because Mercury's anomalous precession can be described in terms of a test particle moving in a Schwarzschild metric. The Schwarzschild metric has a timelike Killing vector, so there is a conserved energy-momentum vector for test particles.
The second sentence is also misleading, since it doesn't make the global/local distinction. What's conserved locally isn't a pseudotensor, it's a tensor (the energy-momentum vector). Globally, there are various pseudotensors that can be defined, and the fact that they're pseudotensors rather than tensors means that they're fundamentally not well-defined quantities -- they require some specially chosen system of coordinates.
Best Answer
Electromagnetic waves (light) are ripples in the electromagnetic field, instead of the spacetime metric.
As far as other forces making waves: Each of the fundamental forces has a particle that carries that force. For electromagnetism, it's the photon, which is the minimum unit of energy in an electromagnetic wave. Similarly, a graviton is the minimum unit of energy in a gravitational wave. These are the force carrier particles for EM and gravity. The W and Z bosons carry the weak nuclear force, but they are too heavy and unstable to last long enough to identify as a wave. Nuclear forces are a little weirder than gravity and EM. I would say they do not have ripples in the same way, but you can read more here: nuclear force waves.