I know TiO2 reflects better than Al2O3 but I can't understand why, in most radiation detection applications using scintillation, aluminum oxide is used as a reflective coating more than titanium dioxide?
The dielectric constant of TiO2 is substantially higher (about 100, depending on a number of things). This leads to a higher refractive index that in turn affects the internal reflection angle, which finally allows making smaller particles that would absorb less light.
Reflection is indeed wavelength dependent and this defines the optimal size of particles for each wavelength. The best average reflection for white light by TiO2 is with the average particle size 0.2-0.3um, 0.15um is the best bor a the blue light. However, particles under 0.1um (nano-particles) represent a serious health risk.
The dependence on the particle size is a smooth function, so 8um particles still would be very reflective, even if not optimal. I've used 44um particles and visually they seemed very reflective. TiO2 is very common, so finding smaller particle sizes should not be too hard. Just keep the health concerns in mind (0.15um average contains plenty of smaller than 0.1um). Do not inhale or allow a skin contact.
Please note that there are two types of TiO2, anatase and rutile with the former better reflecting blue. Please see this link for a lot of information on the subject:
You can see there under Optical theory that the refractive index of TiO2 is superior to most other materials. I don't see there the data on Al2O3 (also known as saphire), but its refractive index is around 1.8, much lower than TiO2. Please see if this interactive page is helpful: