gr.group theory - Is a quotient of a reductive group reductive?

It is important in answering this question that one can extend scalars to a perfect (e.g., algebraically closed) ground field, as was implicit in many of the other answers even if not said explicitly. Indeed, if $k$ is an imperfect field then it always happens that there exist many examples of pairs $(G,H)$ with $G$ a smooth connected affine $k$-group containing no nontrivial smooth connected unipotent normal $k$-subgroup and $H$ a smooth connected normal $k$-subgroup in $G$ such that $G/H$ contains a non-trivial smooth connected unipotent normal $k$-subgroup.
This can even happen when $G$ and $H$ are perfect (i.e., equal to their own derived groups), which is really disorienting if one is accustomed to working with reductive groups.

For a simple commutative example, let $k'/k$ be a purely inseparable extension of degree $p = {rm{char}}(k)$ and let $G$ be the Weil restriction ${rm{Res}} _{k'/k}(mathbf{G} _m)$, which is just "${k'}^{times}$ viewed as a $k$-group". This contains a natural copy of $mathbf{G}_m$, and since $k'/k$ is purely inseparable this $k$-subgroup $H$ is the unique maximal $k$-torus and the quotient $G/H$ is unipotent of dimension $p-1$ (over $overline{k}$ it is a power of $mathbf{G} _a$ via truncation of $log(1+x)$ in degrees $< p$, as one sees using the structure of the $overline{k}$-algebra $overline{k} otimes_k k'$). The main point then is that $G$ itself contains no nontrivial smooth connected unipotent $k$-subgroups, which is true because we are in characteristic $p > 0$ and $G$ is commutative with $G(k_s)[p] = {k'_s}^{times}[p] = 1$! Note: the unipotent quotient $G/H$ is an example of a smooth connected unipotent $k$-group (even commutative and $p$-torsion) which contains no $mathbf{G}_a$ as a $k$-subgroup (proof: commutative extensions of $mathbf{G}_a$ by $mathbf{G}_m$ split over any field, due to the structure of ${rm{Pic}}(mathbf{G}_a)$ and a small calculation); that is, $G/H$ is a "twisted form" of a (nonzero) vector group, which never happens over perfect fields.

Making examples with perfect $G$ and $H$ is less straightforward; see Example 1.6.4 in the book "Pseudo-reductive groups".

As for the suggestion to use Haboush's theorem (whose proof I have never read), I wonder if that is circular; it is hard to imagine getting very far into the theory of reductive groups (certainly to the point of proving Haboush's theorem) without needing to already know that reductivity is preserved under quotients (a fact that is far more elementary than Haboush's theorem, so at the very least it seems like killing a fly with a sledgehammer even if it is not circular).

Finally, since nobody else has mentioned it, look in any textbook on linear algebraic groups (Borel, Springer, etc.) for a proof of the affirmative answer to the original question. For example, 14.11 in Borel's book. Equally important in the theory is that for arbitrary smooth connected affine groups, formation of images also commutes with formation of maximal tori and especially (scheme-theoretic) torus centralizers; see corollaries to 11.14 in Borel's book.