A local ring $(R,\mathfrak{m})$ with $\operatorname{depth} R \ge \mu_R(\mathfrak{m})-1$ is often termed an abstract hypersurface (note this includes the case $\operatorname{depth} R=\mu(\mathfrak{m})$ i.e. when $R$ is regular). Sometimes this condition is defined instead to be that the completion is a hypersurface in the absolute sense, i.e., that $\hat{R} \cong S/(f)$ for some regular local ring $S$ and $f \in S$. That these are equivalent will be shown below. In some sense, this second definition is more natural since one defines an abstract complete intersection in the analogous manner.
We observe the following:
Suppose $R$ is an abstract hypersurface and suppose $R$ is the homomorphic image of a regular local ring. Then $R$ is a hypersurface in the absolute sense, i.e., $R \cong S/(f)$ for some regular local ring $S$ and $f \in S$.
Proof: We may write $R \cong S/I$ where $(S,\mathfrak{n})$ is a regular local ring and $I \subseteq \mathfrak{n}^2$. By assumption, $\operatorname{depth} S-\operatorname{depth} R=\mu_R(\mathfrak{m})-\operatorname{depth} R \le 1$, and thus, by Auslander-Buchsbaum, $\operatorname{pd}_S R \le 1$. If $\operatorname{pd}_S R=0$, then $R \cong S$ and we are done. Otherwise, $\operatorname{pd}_S R=1$. But we have an exact sequence $0 \to I \to S \to R \to 0$. As $\operatorname{pd}_S R=1$, this forces $I$ to be a free $S$-module, and thus $I$ is principal.
In particular, the above theorem shows that the completion of any abstract hypersurface is always a hypersurface in the absolute sense, since $\mu_R(\mathfrak{m})$ and $\operatorname{depth} R$ are preserved by completion, and thus in turns shows that abstract hypersurfaces enjoy many of the same homological properties enjoyed by hypersurfaces in the absolute sense; in particular they are always Gorenstein.
To answer your question directly, no; an example was given by Heitmann and Jorgensen of an abstract hypersurface which is not a hypersurface in the absolute sense. See the (excellently titled) paper ``Are complete intersections complete intersections?" https://arxiv.org/pdf/1109.4921.pdf.
Your argument is very useful. I follow your argument. Since $R/I$ has finite length, we know $$R/I\cong \mathrm{Hom}_R(\mathrm{Hom}_R(R/I,E_R(k)),E_R(k))=\mathrm{Hom}_R(N,E_R(k)).$$ You have showed that $\mathrm{pd}_RN<\infty$, so we know $\mathrm{id}_R(R/I)<\infty$.
Lemma 1. Let $(R,m,k)$ be a Noetherian local ring. If $M$ is a finitely generated $R$-module, then $\mathrm{id}_RM=\sup\{i\mid \mathrm{Ext}^i_R(k,M)\neq 0\}$.
By lemma 1, and by Nakayama we know:
Lemma 2. Let $(R,m,k)$ be a Noetherian local ring, $M$ is a finitely generated $R$-module, and $x\in m$ is a regular element on $M$. Then $\mathrm{id}_R(M)=\mathrm{id}_R(M/xM)$.
So we know $\mathrm{id}_R(R)=\mathrm{id}_R(R/I)<\infty$, since $I$ is generated by a regular sequence.
More generally, Foxby showed:
If $R$ is a Noetherian local ring, if there exists a finitely generated module with finite injective dimension and finite projective dimension, then $R$ is Gorenstein.
Best Answer
If $\mathfrak p\in \mathrm{Ass}(M)$, then $\dim(R/\mathfrak p)=\mathrm{depth}(M)$, so $\dim(R/\mathfrak p)=1$. It follows that $\mathrm{ht}(\mathfrak p)=1$. From Auslander-Buchsbaum we get $\mathrm{pd}_{R_{\mathfrak p}}(M_{\mathfrak p})+\mathrm{depth}(M_{\mathfrak p})=1$. Since $\mathfrak p\in \mathrm{Ass}(M)$ we have that $\mathrm{depth}(M_{\mathfrak p})=0$, and thus $\mathrm{pd}_{R_{\mathfrak p}}(M_{\mathfrak p})=1$, so $M_{\mathfrak p}$ is not free.