Yes, there is.
Lemma 1. Let $G$ be any $n$-vertex connected graph in which $\Omega(n)$ vertices have degree 3 or more. Then $G$ contains a connected dominating set $D$ whose complement has size $\Omega(n)$.
Before we give the proof, note that this answer strengthens this result, showing that it suffices for
$G$ to meet the following condition: the number of edges in $G$ such that both endpoints have degree at most two is $O(1)$ times the number of vertices in $G$ of degree three or more.
Proof of Lemma 1.
Consider any spanning tree $T$ of $G$. Let $n_1$ and $n_2$ be the numbers of leaves and degree-2 vertices in $T$. Let $n^+_3$ be the number of vertices of degree three or more in $T$.
Let $G_2$ be the subgraph of $T$ consisting of the $n_2$ vertices of degree 2 in $T$ and the edges in $T$ between these vertices. (Note that $G_2$ is a subgraph of $T$, and has maximum degree 2, so is a collection of paths.) We claim that the number of edges in $G_2$ is at least $n - 4n_1$, and the number of vertices in $G_2$ is at least $n-2n_1$.
To show the claim, consider the tree $T'$ obtained from $T$ by splicing out all degree-2 vertices (that is, for each maximal path $v_1, v_2, \ldots, v_p$ in $T$ such that each intermediate vertex $v_i$ for $2\le i \le p-1$ has degree two in $T$, remove the path edges and the intermediate vertices, replacing them all by the edge $(v_1, v_p)$, which might not be in $G$). It is well known (and easy to show) that in $T'$ (as in any tree where each non-leaf node has degree 3 or more) the number of edges is at most twice the number of leaf nodes, i.e., at most $2n_1$. And for every edge in $T'$, $G_2$ is missing at most two edges from $T$, so $G_2$ is missing at most $4n_1$ edges of $T$. Likewise, $T'$ has at most $2n_1$ non-leaf nodes, to $G_2$ is missing at most $2n_1$ nodes of $T$. This shows the claim.
It follows that $G_2$ has at most $4n_1$ more nodes than edges, so consists of at most $4n_1$ paths (counting each isolated vertex in $G_2$ as a path). It follows that at least $n-10 n_1$ nodes in $G_2$ have two neighbors in $G_2$. That is, back in $T$, at least $n-10 n_1$ nodes in $T$ have the following property: they have degree 2 in $T$, and both of their neighbors have degree 2 in $T$. Call such a node typical.
Now fix $T$ to be a spanning tree in $G$ with a maximum number of leaves.
Let $u$ be a typical vertex in $G$ and consider any edge $(u,w)$ from $u$ that is not in $T$. Consider adding the edge $(u, w)$ to $T$, creating a cycle, and removing one of the other edges incident to $u$ to break the cycle (so the result is again a spanning tree). This makes one of $u$'s neighbors into a leaf, so by the choice of $T$ the vertex $w$ must have been a leaf in $T$ (otherwise the operation would increase the number of leaves). So, for every typical vertex $u$, each edge from $u$ except the two in $T$ goes to a leaf in $T$.
Assume for contradiction that $n_1 = o(n)$.
Then there are at least $n - 10 n_1 = n - o(n)$ typical nodes in $G$, and at least $\Omega(n)$ nodes in $G$ have degree at least three. It follows that at least one of the $n_1$ leaves, say $u$, has two edges to typical vertices, say $v_1$ and $v_2$. But then by adding edges $(u, v_1)$ and $(u, v_2)$ to $T$, and removing one edge from $v_1$ and one from $v_2$ (thereby making a neighbor of each a leaf in the new tree), we can create a spanning tree with more leaves than $T$, contradicting the choice of $T$. So $n_1 = \Omega(n)$.
The non-leaf vertices of $T$ form a connected dominating set, whose complement (the leaves) has size $n_1 = \Omega(n)$. $~~~~~\Box$