If $G$ is nonabelian simple, then $H$ cannot exist.
Lemma : Let $\cal X$ be a collection of normal subgroups of a group $H$ such that $\bigcap \cal X = 1$, and assume that $H/X$ is nonabelian simple for all $X \in \cal X$. Then $H$ is isomorphic to a direct product of nonabelian simple groups.
Proof: Let $M \triangleleft H$ be minimal such that $H/M$ is isomorphic to a direct product of nonabelian simple groups. To complete the proof, we show $M = 1$. Otherwise, there exists $X \in \cal X$ such that $M \cap X < M$, and we write $D = M \cap X$. Now
$M \not\subseteq X$, so $MX > X$, and since $H/X$ is simple, it follows that $MX = G$. Then $H/D \cong M/D \times X/D$, and this is a direct product of nonabelian simple groups since $M/D \cong H/X$ is nonabelian simple and $X/D \cong H/M$ is a direct product of nonabelian simples. This contradicts the minimality of $M$.QED
Theorem: A finite group $H$ cannot be covered by a collection $\cal X$ of normal subgroups such that $H/X$ is nonabelian simple for all $X \in \cal X$.
Proof: Assume $H$ actually can be so covered, and let $D = \bigcap \cal X$. Then $D \triangleleft H$ and $H/D$ is covered by the normal subgroups $X/D$, and each of these has a nonabelian simple quotient. We can thus replace $H$ by $H/D$, so we can assume $D = 1$. The lemma then applies, and $H$ is a direct product of a collection $\cal Y$ of nonabelian simple subgroups. Then every normal subgroup of $H$ is a product of some subset of $\cal Y$, and proper normal subgroups are products of proper subsets of $\cal Y$. Now choose one nonidentity element of each member of $\cal Y$ and let $h$ be the product of the chosen elements. Then $h$ lies in no proper normal subgroup of $H$, and this is a contradiction.