Does replacing each vertex of $G$ by $H$ increase treewidth of $G$ by at most $\Delta(G)$?
Let $G$ be a complete graph with $n$ vertices and $H$ be a complete graph with $n-1$ vertices, all marked. We replace each vertex of $G$ with $H$ to produce a graph $G^*$ with $n(n-1)$ vertices that is $n$-regular. We claim that the treewidth of $G^*$ is at least $n (n-1)/12$, which falsifies the conjecture. Proof:
Suppose that the treewidth of $G^*$ is less than $n(n-1)/12$. Therefore by standard results [1] its vertices can be partitioned to sets $A$, $S$, $B$ such that there are no edges between $A$ and $B$, and $|S| = n(n-1)/12$, $|A| \ge n(n-1)/4$, $|B| \ge n(n-1)/4$, and $|A| + |B| = n(n-1)11/12$.
We call the cliques $H$ in $G^*$ corresponding to vertices of $G$ clusters. The graph $G^*$ has $n$ clusters. Note that each cluster has vertices from at most one of $A$ or $B$. Let $A_f$ be the fraction such that $n A_f$ clusters contain vertices from $A$ and $B_f$ the fraction such that $n B_f$ clusters contain vertices from $B$.
Now, $A_f \ge 1/4$, $B_f \ge 1/4$, and $A_f + B_f \ge 11/12$.
Notice that if there is a cluster $v$ that contains vertices from $A$ and a cluster $u$ that contains vertices from $B$ then these clusters are connected by an edge $\{v_i, u_j\}$, both vertices in it unique to that pair of clusters, and at least one of $v_i$ and $u_j$ should be in $S$.
Therefore we get that $|S| \ge n^2 A_f B_f$, which we can solve with our constraints to a minimum value $|S| \ge n^2 A_f B_f \ge n^2 8/12 \cdot 1/4 = n^2 1/6$, which is a contradiction to the assumed size of $S$.
[1] http://parameterized-algorithms.mimuw.edu.pl/parameterized-algorithms.pdf