Over a (commutative) local ring (non necessarily noetherian), any finitely generated flat module is free (Matsumura, Commutative Algebra, Prop. 3.G, p. 21), hence projective.
According to Bourbaki, Commutative Algebra, this is false for general rings (page 29, Example (2)). See here.
The counterexample is the following: take the direct product $A$ of the rings $\mathbb Z/p$ over all primes, and consider the ideal $I$ obtained from taking the direct sum. Then $I$ is projective and faithful, but not faithfully flat.
To see that $I$ is not faithfully flat, recall that a module $M$ is faithfully flat iff it is flat and $JM\neq M$ for every (maximal) ideal. But note that $I^2=I$, so it follows that $I$ is not faithfully flat.
On the other hand, if we take $e_p\in I$ the $p$-th coordinate element, we see that for $a\in A$, $ae_p = a_p$, so if $a$ annihilates $M$ we deduce that $a=0$, and $M$ is faithful.
Finally, $I$ is projective: I will exhibit a dual basis for it. Consider the coordinate elements $e_p$ and the $A$-linear maps $e_p^*$ that are the composition of the projection to the $p$th component with the inclusion $\mathbb Z/p\to A$. It is immediate from the definitions that if $x\in I$, then $\sum_p e_p^*(x)e_p =x$
Best Answer
Yes, this is true.
Indeed, we can prove the following.
First, $M\otimes_R S$ is non-zero as $\pi$ is faithfully flat. For each prime ideal $\mathfrak{q}$ of $S$. We denote $\mathfrak{p}=\pi^{-1}\mathfrak{q}$. This is a prime ideal of $R$. Note that $(M\otimes_R S)_{\mathfrak{q}}\cong M\otimes_R S_{\mathfrak q}\cong M_{\mathfrak{p}}\otimes_{R_{\mathfrak p}}S_{\mathfrak q}$. This is non-zero. The point is the induced map $R_{\mathfrak p}\rightarrow S_{\mathfrak q}$ is faithfully flat and $M_{\mathfrak p}\neq 0$ as $\text{Supp}_R(M)=\text{Spec}(R)$. To see $R_{\mathfrak p}\rightarrow S_{\mathfrak q}$ is faithfully flat, we just need to note that the flat homomorphism of local rings is faithfully flat.