If $R$ is not a local ring can we say that $R_p$ is not faithfully flat as an $R$- module for any prime ideal $p$ of $R$? If $R$ is not a local ring can we say that $R_p$ is not faithfully flat as an $R$ module for any prime ideal $p$ of $R$?
For local ring since the natural map $R \to R_p$ is a homomorphisms of local rings and $R_p$ is a flat $R$-module, hence $R_p$ is faithfully flat $R$-module. But in the above situation (non-local ring) how can I approach ?
 A: a) For every prime ideal $\mathfrak p \subset \mathbb Z$ the morphism $\mathbb Z\to \mathbb Z_\mathfrak p$ is not faithfully flat since the dual map map $$\operatorname {Spec}\mathbb Z_\mathfrak p \to \operatorname {Spec}\mathbb Z$$ is not surjective.
b) By the way, the statement in your second sentence is false: the map $R\to R_\mathfrak p$ is in general not faithfully flat, even for a local ring $R$.
For example if $R$ is the local ring $R=\mathbb C[X,Y]_{\langle X,Y\rangle}$ the morphism $R\to R_{\langle 0\rangle}=\mathbb C(X,Y)$ is not faithfully flat since the dual map $$\operatorname {Spec}\mathbb C(X,Y) =\{*\}\to \operatorname {Spec}R$$ with source a singleton set is certainly not surjective.
A: Recall that a faithfully flat epimorphism is always an isomorphism [Stacks 04VU].  
A localization is always a flat epimorphism.  If additionally it is faithfully flat, then the localization is an isomorphism, and you can check that this forces the localizing set to consist only of units.  In particular if $R_P$ is faithfully flat for a prime $P$, then $P$ is the unique maximal ideal of $R$.  
