To answer (in one case) the particular question it is always true for linear algebraic groups that the orbit space of a closed subgroup is a scheme, and it will also be a linear algebraic group provided that the subgroup is normal.
Suppose that we have a closed subgroup H of G. One way to build the orbit space is as a G-orbit in the projective bundle P(V) associated to some rational representation G -> GL(V). The point is that we can pick such a representation having the property that V has a one dimensional subspace W so that H is precisely the subgroup of G fixing W. Taking [W] in the associated projective space of V we get that H is the isotropy group of [W]. One can then identify G/H with the orbit of [W] in P(V) and it can be checked that this is the right thing to do in terms of the universal property. In particular the orbit space is quasi-projective and one can go onto study when the orbit space is projective (i.e. look at parabolic subgroups).
I am not sure about the first part... If the action is transitive often one can argue that the quotient is representable by an algebraic space and then use transitivity of the group action to show that one in fact has a scheme. I am pretty sure freeness should correspond to good properties of the geometric quotient provided one exists (probably only then on some open subscheme) rather than guaranteeing existence and it seems that this is the case given the example Charles linked to.