You should basically never expect anything to have a subobject classifier -- it's a ridiculously strong condition. Some examples showing how quickly it fails to exist:
The category of groups is complete. If this category had a subobject classifier Ω, every subgroup would be the kernel of a map to Ω. But not every subgroup is normal.
The category of Hausdorff spaces is complete. If it had a subobject classifier Ω, every subspace would be the preimage of a point in Ω. But not every subspace is closed.
The category of rings with identity is complete. The terminal object is the zero ring (with 0 = 1). But the zero ring has no maps to any nonzero ring.
The category of compact Hausdorff spaces is complete. As above every closed subspace should be the preimage of a point $\ast$ in Ω under a unique map to Ω. For the subspace {1} in {1,2} to be classified by a unique map, $\Omega\setminus \ast$ must be a single point, so Ω is discrete on two points. But then there is only one map from $S^1$ to Ω, while it has uncountably many subobjects.
Edit: the opposite of any cocomplete category will (with probability 1) give examples as well.
The category of groups is cocomplete. A subobject classifier in the opposite category would be a group Ω so that every surjection G -> H is a pushout of Ω -> 1 and Ω -> G. This is the amalgamated free product of G with the trivial group over Ω, that is the quotient of G by the image of Ω. But if Ω is a group surjecting to the kernel of every map, that surjection cannot be unique (just consider Z -> 1).
The category of sets (also Hausdorff spaces) is cocomplete. In the opposite category Ω would be a set (space) so that every surjection X -> Y was the quotient of X by the image of a map Ω -> X; this would imply that at most one fiber of X -> Y is not just a single point.
The category of commutative rings with identity is cocomplete. A subobject classifier for the opposite category (the category of affine schemes) would be an affine scheme Ω with a $\mathbb{Z}$-point Spec $\mathbb{Z}$ -> Ω so that any monomorphism Y -> X is the base change to $\mathbb{Z}$ of a unique map X -> Ω. This is ridiculous; note that Spec $\mathbb{Q}$ -> Spec $\mathbb{Z}$ is injective, and you'll never get Spec $\mathbb{Q}$ as the fiber product of two $\mathbb{Z}$-points (and in general Y need not have any $\mathbb{Z}$-points).\mathbb{Z}$-points.
The partial order category of ordinals is cocomplete. The initial object is 0; but no nonzero ordinal maps to (is ≤) 0. Thus Ω would have to be 0, but then we would have for any α ≥ β that α was the coproduct (supremum) of β with 0, which is false whenever α > β.
