In positive characteristic, the normalizer of a (connected) reductive subgroup of a (connected) reductive group is not in general reductive.
I communicated the example below in an emailed answer to a query in April 2002 (it took me some searching in old emails to find it!) At the time I wrote something at the end like "I'm not aware of a good reference" -- this remains true.
Let $k$ be alg. closed of positive characteristic. I will exhibit $H < G$ two reductive groups such that $C_G(H)$ is not reductive; since $C_G(H)$ is normal in $N_G(H)$, also $N_G(H)$ is not reductive.
A slight modification of the example also shows that the group $C_G(H)/Z(H)$ can be unipotent (here $Z(H)$ is the center of $H$).
Here is the example. Let $n \ge 2$, and consider the adjoint representation $$\operatorname{Ad}: \operatorname{GL}(n,k) \to \operatorname{GL}(L)$$ where $L$ is the Lie algebra of $\operatorname{GL}(n,k)$.
The image $H$ of $\operatorname{Ad}$ is a reductive subgroup of $G=\operatorname{GL}(L)$, and the centralizer $C_G(H)$ identifies with the group of units of the endomorphism ring $\operatorname{End}_H(L)$.
If $(n,p) = 1$, $L$ is a semisimple representation of $\operatorname{GL}(n,k)$ ith two distinct irreducible factors, so $\operatorname{End}_H(L) = k \times k$, and in this case $C_G(H)$ is therefore a 2 dimensional torus hence is reductive.
If $(n,p) = p$, $L$ is an indecomposable representation with 3 composition factors. Thus the endomorphism ring of $L$ is a local ring. It turns out that this endomorphism ring $\operatorname{End}_H(L)$ still has dimension 2, but it is now isomorphic to the ring $k[t]/(t^2)$. To exhibit a non-0 nilpotent $H$-endomorphism $t$ of $L$, take the following map: to a matrix $X$ in $L$, assign the matrix $t(X) = \operatorname{tr}(X).1$ where 1 denotes the $n\times n$ identity matrix, and $\operatorname{tr}()$ denotes the trace. Since $n$ is divisible by $p$, applying $t$ twice gives 0.
The unit group of this local ring is isomorphic to the product of a $1$ dimensional torus and the additive group of the field; such a group is not reductive. The additive subgroup is precisely the set of all automorphisms $1 + at$ of $L$ with $a \in k$.
Now restrict $\operatorname{Ad}$ to $\operatorname{SL}(n,k)$; this is not the adjoint representation of $\operatorname{SL}(n,k)$ as we do not change $L$. Let $H'$ be the image of $\operatorname{SL}(n,k)$ in $G'=\operatorname{SL}(L)$. Then $C_{G'}(H')$ is isomorphic to the additive group of the field and is thus unipotent.