You don't have to orbit, you can just use a rocket to stay put. All observers that can communicate with infinity for all time agree about the infalling object. It gets frozen and redshifted at the horizon.
EDIT: in response to question
The issue of two objects falling in one after the other is adressed in this question: How does this thought experiment not rule out black holes? . The answers there are all wrong, except mine (this is not an arrogant statement, but a statement of an unfortunate fact).
When you are near a black hole, in order to stay in place, you need to accelerate away from the black hole. If you don't, you fall in. Whenever you accelerate, even in empty Minkowksi space, you see an acceleration event horizon behind you in the direction opposite your acceleration vector. This horizon is a big black wall that follows you around, and you can attribute the various effects you see in the accelerating frame, like the uniform gravitational field and the Unruh radiation, to this black-wall horizon that follows you around.
When you are very near a black hole, staying put, your acceleration horizon coincides with the event horizon, and there is no way to tell them apart locally. This is the equivalence principle, in the form that it takes in the region by the horizon where there is no significant curvature.
The near-horizon Rindler form of the metric allows you to translate any experiment you can do in the frame near a black hole to a flat space with an accelerating observer. So if you measure the local Hawking temperature, it coincides with the Unruh temperature. If you see an object fall and get redshifted, you would see the same thing in empty space, when accelerating.
The point is that the acceleration you need to avoid falling in is only determined globally, from the condition that you stay in communication with infinity. If you stop accelerating so that you see the particle cross the horizon, the moment you see the particle past the horizon, you've crossed yourself.
Everything that passes the event horizon of a black hole falls into the singularity, including photons. That's why it's a singularity.
There is a particular radius outside the event horizon where a photon will orbit, but the orbit is unstable- if the photon gets perturbed a little closer or the black hole's mass increases at all, it will fall in, and if the black hole's mass decreases due to Hawking radiation or the photon gets perturbed away from the black hole, it will escape.
Best Answer
Whether the infalling material is matter or antimatter makes no difference.
Fundamentally, the confusion probably comes from thinking of black holes as normal substances (and thus retaining the properties of whatever matter went into making them). Really, a black hole is a region of spacetime with certain properties, notably the one-way surface we call an event horizon. That's it. Whatever you envision happening on the inside of a black hole, whether it be a singularity or angels dancing on the head of a pin, is completely irrelevant.
The reason spacetime is curved enough to form an event horizon is essentially the due to the density of mass and energy in the area. Antimatter counts just the same as matter when it comes to mass and energy. Anti-protons have the same, positive mass as normal protons, and at a given speed they have the same, positive kinetic energy too.
Even if you wanted matter and antimatter to annihilate somewhere near/inside a black hole, the resulting photons would cause no less curvature of spacetime, as all particle physics reactions conserve energy and momentum. This is related to how you could form a black hole from nothing but radiation.