No. Or rather, it depends on which diagonal you're talking about. One will allow |H>+|V> through and completely block |H>-|V> ; the other will do the reverse. You can pretty much see what happens here just by looking at things: the lower-left to upper-right diagonal clearly corresponds to |H>+|V> -- that line is given by [tex]c (\hat{x}+\hat{y}) ,\ c \in R[/tex] after all -- and so a polarizer along that lines corresponds to a projection onto |H>+|V>. The intensity of light getting through is given by [tex]| <\mbox{light}|\mbox{polarizer}>|^2[/tex] where light is the normalized polarization vector of the light, and polarizer is the normalized vector of the polarizer.
In a more general situation -- I mean when you're dealing with polarization of objects which are not spin-1 like light -- what you'd want to do is rotate the object's polarization vector into the basis of the polarizer. To do this you need to know how the object transforms under rotations; that can be found by exponentiating the angular momentum operators (which are the generators of rotations.) If this sounds like gobbedy-gook, you can study a good (grad-level) quantum mechanics textbooks. They will go into this in some depth. IIRC Sakurai's Modern Quantum Mechanics is particularly good on this topic. For a spin-m object polarized perpendicularly to the plane of the polarizer, this turns out to be pretty simple; you get roughly the same result as for light, except the rotation angle that shows up is proportional to m*theta, ie
[tex]|\mbox{rotated ket}> = \left( \begin{array}{cc} \cos(m\theta) & \sin(m\theta) \\ -\sin(m\theta) & \cos(m\theta) \end{array} \right) |\mbox{original ket}>[/tex]
