Emil Jeřábek gave more complete answer than mine, so I have initially deleted it. On the afterthought, I decided to turn it into an addendum to Emil's answer. It is a generalization of sorts: if an embedding of a complete Heyting algebra $H$ into a co-Heyting algebra preserves finite joins and arbitrary meets (not necessarily infinite joins or implication), then $H$ itself is co-Heyting.

Indeed given such an embedding $i$, any $a\in H$ and any $S\subseteq H$ will satisfy $$ i(a\lor\bigwedge S)=i(a)\lor i(\bigwedge S)=i(a)\lor\bigwedge i(S)=\bigwedge(i(a)\lor i(S))=\bigwedge i(a\lor S)=i(\bigwedge(a\lor S)), $$ hence $a\lor\bigwedge S=\bigwedge(a\lor S)$.

Thus in particular if $H$ is not co-Heyting (and, for example, algebras of open sets of topological spaces usually are not), then any embedding of $H$ into some $\mathsf{Up}(X)$ fails to preserve some meet.

Emil Jeřábek gave more complete answer than mine, so I have initially deleted it. On the afterthought, I decided to turn it into an addendum to Emil's answer. It is a generalization of sorts: if an embedding of a complete Heyting algebra $H$ into a co-Heyting algebra preserves finite joins and arbitrary meets (not necessarily infinite joins or implication), then $H$ itself is co-Heyting.

Indeed given such an embedding $i$, any $a\in H$ and any $S\subseteq H$ will satisfy $$ i(a\lor\bigwedge S)=i(a)\lor i(\bigwedge S)=i(a)\lor\bigwedge i(S)=\bigwedge(i(a)\lor i(S))=\bigwedge i(a\lor S)=i(\bigwedge(a\lor S)), $$ hence $a\lor\bigwedge S=\bigwedge(a\lor S)$.

Thus in particular if $H$ is not co-Heyting (and, for example, algebras of open sets of topological spaces usually are not), then any embedding of $H$ into any $\mathsf{Up}(X)$ fails to preserve some meet.

Every $\mathsf{Up}(X)$ satisfies $U\cup\bigcap\mathscr U=\bigcap(U\cup\mathscr U)$ for any $U\in\mathsf{Up}(X)$ and any $\mathscr U\subseteq\mathsf{Up}(X)$. So if $H$ admits a complete embedding $i$ (even just an embedding preserving arbitrary meets) to some $\mathsf{Up}(X)$, the same will hold in $H$: for any $a\in H$ and any $S\subseteq H$ one will have $$ i(a\lor\bigwedge S)=i(a)\cup i(\bigwedge S)=i(a)\cup\bigcap i(S)=\bigcap(i(a)\cup i(S))=\bigcap i(a\lor S)=i(\bigwedge(a\lor S)), $$ hence $a\lor\bigwedge S=\bigwedge(a\lor S)$.

Thus if $H$ is not co-Heyting (and, for example, algebras of open sets of topological spaces usually are not), it cannot be completely embedded in any $\mathsf{Up}(X)$.

Emil Jeřábek gave more complete answer than mine, so I have initially deleted it. On the afterthought, I decided to turn it into an addendum to Emil's answer. It is a generalization of sorts: if an embedding of a complete Heyting algebra $H$ into a co-Heyting algebra preserves finite joins and arbitrary meets (not necessarily infinite joins or implication), then $H$ itself is co-Heyting.

Indeed given such an embedding $i$, any $a\in H$ and any $S\subseteq H$ will satisfy $$ i(a\lor\bigwedge S)=i(a)\lor i(\bigwedge S)=i(a)\lor\bigwedge i(S)=\bigwedge(i(a)\lor i(S))=\bigwedge i(a\lor S)=i(\bigwedge(a\lor S)), $$ hence $a\lor\bigwedge S=\bigwedge(a\lor S)$.

Thus in particular if $H$ is not co-Heyting (and, for example, algebras of open sets of topological spaces usually are not), then any embedding of $H$ into some $\mathsf{Up}(X)$ fails to preserve some meet.