If the sun were a true black body, then it would be black. The extent to which it differs from a true black body can be seen from the sun-light spectrum outside the earth's atmosphere. Overall, it is pretty close to a 5778 K black body, so the sun will likely be very dark if you remove its own thermal radiation. Upon closer inspection, you can see that it lacks a bit in the UV and it has excessive radiation around blue-green visible light (500 nm). The explanation for the excess can be found in "How can it be that the sun emits more than a black body?": some short-wavelength light from deeper, hotter layers penetrates through the "cooler" top layer.
So, the outermost layers of the sun are either partially transparent, or black. They do not reflect. Remember that for everyday objects to reflect, they need to involve abrupt changes of the optical properties. Paper and sugar consist of transparent, colorless material; it appears white because of the reflections at the grain boundaries. The sun does not have a well defined surface; rather, there is a very gradual transition from interstellar vacuum to the more dense parts of the sun.
Update after comment The black-body deficit in the UV could be the result of Rayleigh scattering, which would suggest that the sun may be a bit bluish. Only a fraction 4e-4 of the hydrogen in the photosphere is ionized (source: What is the degree of ionization is the solar photosphere?); I would expect that molecular or atomic hydrogen has a very small Rayleigh cross section compared to the earth's atmosphere. I'm not sure what the small plasma fraction in the photosphere would do, exactly. In any case, if the cross sections for inelastic scattering (absorption and thermal emission) are much larger than the cross sections for elastic scattering (Rayleigh scattering), then you won't see much of the elastic scattering, just like adding a drop of milk to black ink will not give the ink a gray appearance.