I think this is certainly true if $p$ is generated by a regular sequence. In this case, $R/p$ is regular, local and hence Cohen Macaulay. Hence, $R/p$ is a maximal Cohen Macaulay module over $R/Ann(R/p)$. But $Ann(R/p)=p$ since $p$ is prime. So, $R/p$ is a maximal Cohen Macaulay $R/p$-module.

Essentially, all you need is for $R/p$ to be Cohen Macaulay. But probably weaker conditions might suffice (Edit: Also, $R/p$ is Cohen Macaulay when $R$ is regular, local, iff $p$ has height $1$)

I think this is certainly true if $p$ is generated by a regular sequence. In this case, $R/p$ is regular, local and hence Cohen Macaulay. Hence, $R/p$ is a maximal Cohen Macaulay module over $R/Ann(R/p)$. But $Ann(R/p)=p$ since $p$ is prime. So, $R/p$ is a maximal Cohen Macaulay $R/p$-module.

Essentially, all you need is for $R/p$ to be Cohen Macaulay. But probably weaker conditions might suffice (Edit: Also, $R/p$ is Cohen Macaulay when $R$ is regular, local, iff $p$ has height $1$)

The question is similar to small Cohen-Macaulay module conjecture where we ask the same question over a complete local ring (which I believe is open)

I think this is certainly true if $p$ is generated by a regular sequence. In this case, $R/p$ is regular, local and hence Cohen Macaulay. Hence, $R/p$ is a maximal Cohen Macaulay module over $R/Ann(R/p)$. But $Ann(R/p)=p$ since $p$ is prime. So, $R/p$ is a maximal Cohen Macaulay $R/p$-module.

Essentially, all you need is for $R/p$ to be Cohen Macaulay. But probably weaker conditions might suffice.

I think this is certainly true if $p$ is generated by a regular sequence. In this case, $R/p$ is regular, local and hence Cohen Macaulay. Hence, $R/p$ is a maximal Cohen Macaulay module over $R/Ann(R/p)$. But $Ann(R/p)=p$ since $p$ is prime. So, $R/p$ is a maximal Cohen Macaulay $R/p$-module.

Essentially, all you need is for $R/p$ to be Cohen Macaulay. But probably weaker conditions might suffice (Edit: Also, $R/p$ is Cohen Macaulay when $R$ is regular, local, iff $p$ has height $1$)