The graph from here, found via a House of Graphs search and also pictured below, is $4$-regular, connected, and class $1$ (as the edge coloring shows), but not Hamiltonian:
Also, $3$-regular graph formed by the orange, black, and purple edges in the coloring above is connected, so this means that the answer is "no" to both of your questions.
Firstly, many thanks to Alex K. I have consulted a literature and found that when a $3$-regular graph is connected, this bound can be improved to $\frac{4n-1}{9}$. And the bound is tight.
- [1] Biedl, T., Demaine, E.D., Duncan, C.A., Fleischer, R., Kobourov, S.G.: Tight bounds on maximal and maximum matchings. Discrete Math. 285, 7–15 (2004)
Theorem 13. [1] Any 3-regular connected graph of order $n$ has a matching of size $\frac{4 n-1}{9}$.
Note that $\frac{4 n-1}{9}-\frac{7}{16}n=\frac{n}{144}-\frac{1}{9}$. A $36$-order 3-regular connected graph has a maximum matching with size at least $15$.
Proof. Let $G$ be a 3-regular graph of order $n$. By Theorem 11 and Lemma $12, G$ has a matching of size
$$
\frac{3 n-2 \ell_2}{6} \geqslant \frac{3 n-\frac{n+2}{3}}{6}=\frac{8 n-2}{18}=\frac{4 n-1}{9} .
$$
Theorem 11. Any 3-regular graph $G$ of order $n$ has a matching of
size $\frac{3 n-2 \ell_2}{6}$, where $\ell_2$ is the number of leaves
in the 2-block tree of $G$.
Lemma 12. The 2-block tree of any
3-regular graph of order $n$ has at most $\frac{n+2}{6}$ leaves.
The bound in Theorem 13 is tight, which can be seen by attaching the smallest possible 3-regular graph to every leaf of the graph as follows.
The resulting graph, shown in the following picture, is defined for $n \equiv 16$ mod 18. There are $\frac{n-7}{9}$ black vertices, inducing $\frac{4 n-10}{18}$ odd components. Hence, any matching has at least $\frac{4 n-10}{18}-\frac{n-7}{9}=\frac{n+2}{9}$ unmatched vertices and therefore at most $\frac{8 n-2}{9}$ matched vertices.
Best Answer
Here is a counterexample:
We've got a $4$-regular, Hamiltonian graph on $12$ vertices. However, deleting the highlighted edge leaves two components on $5$ vertices, which don't have a perfect matching, so the highlighted edge is not part of any matching.