I was thinking about homotopy theory and classical questions such as "do homotopy groups characterize the homotopy type of a space ?", and came up with this argument :
$hTop$ (the category of topological spaces and continuous maps up to homotopy) is not concrete, but $Set^\mathbb{N}$ ($\mathbb{N}$ discrete category) is and so $Grp^\mathbb{N}$ as well, hence $(\pi_k)$ cannot be a faithful functor : there are $f, g : X\to Y$ that are not homotopic but such that for all $k$, the induced maps $f_*, g_* : \pi_k(X) \to \pi_k(Y)$ are equal.
I was wondering if in this special case, not faithful implied not conservative : could one deduce from this argument that there are non homotopy-equivalent spaces $X$ and $Y$ that have isomorphic $\pi_k$'s for all $k$ ?
The most natural condition for a conservative functor $F$ to be faithful is that $F$'s domain should have equalizers or coequalizers which $F$ preserves. This condition isn't available in homotopy theory. There's a similar condition involving homotopy (co)equalizers, but for that you need more than just the homotopy category.
As to proving that $(\pi_k)$ is not conservative, well, the first problem is that it's not a functor, as you need to choose basepoints. If you take pointed spacee $hTop_*$, i.e. your category but with based maps and homotopies, then there are two different easy way to see $(\pi_k)$ is not conservative: there are weak equivalences which are not homotopy equivalences, and the homotopy groups see nothing about connected components away from the base point. Thus one should really consider $whTop_{*,x}$, the category of based, connected spaces with weak equivalences inverted. And in that case $(\pi_k)$ is indeed conservative, which is Whitehead's theorem.
If you wanted to stick to the unpointed case, then you could use the functor $\prod_n whTop(S^n,X)$, which is less obviously nonconservative-one trick is to consider $X$ and $Y$ as classifying spaces of groups and note that $whTop(S^1,BG)$ is the set of conjugacy classes of elements of $G$. But in any case, one has to do some kind of a direct proof, unrelated to faithfulness. And there can be no general abstract argument that $whTop$ (or $hTop,hTop_*$, etc) admits no conservative functor into $Sets$, as with the result that homotopy is not concrete, as every locally small category admits such a functor.