See Bourbaki, Groupes et algèbres de Lie, IV.2.7,  Corollary of Theorem 5 (due to Tits).
 We assume that $K$ has at least four elements and that the Lie algebra is absolutely simple. We assume also that $G$ is quasi-split, which is the case when $K$ is an algebraically closed field or a finite field.  We construct a $BN$-pair taking for $B$ a Borel subgroup. Since the root system is irreducible, the Coxeter system is irreducible. If we denote by $G(K)^+$ the (normal) subgroup of $G(K)$ generated by the unipotent elements, Corollary of Theorem 5 says that  $G(K)^+/(G(K)^+\cap Z(K))$ is either a nonabelian simple group or reduces to one element. This proves that then the group $G(K)^+$  has no nontrivial noncentral normal subgroups. When $G$ is  quasi-split and *simply connected*,  $G(K)^+=G(K)$ (Steinberg).
Thus $G(K)$ has no noncentral normal subgroups, which seems to answer the question over an algebraically closed field and over a finite field.