Yes, this is true. See this Math Overflow question for a precise statement and a reference to its proof in Bourbaki's Commutative Algebra.
This result is also stated in my commutative algebra notes, but the proof is not unfortunately not yet written up there. I certainly hope that this will be remedied soon though, as I will be teaching a course out of these notes starting on Monday. When the proof gets written, I will update this answer with a page number.
Added: Here is something in the MO answer that I decided was worth a comment here. For finitely generated modules, this stronger version of local freeness is actually equivalent to projectivity, whereas the weaker "pointwise local freeness" is subtly weaker in general.
For the case $R$ is a local ring it's a corollary of Nakayama's lemma.
As the notation in the above link, suppose $M$ is a finite generated projective module over $R$, then, first pick a minimal number of generators, i.e., $M=Rm_1+\cdots +Rm_k$, and $k$ is the minimal number with this property, so we get a decomposition
$$R^k=M\oplus N,$$ then, we are left to prove $N=0$.
First, applying $R/I\otimes-$, where $I$ is the unique maximal ideal in $R$, then we get $$(R/I)^k=M/IM\oplus N/IN,$$ and note that $M/IM$, $N/IN$ are vector spaces over the field $R/I$, so by comparing the dimension, we get $N/IN=0$, i.e., $N=IN$, then,
we use the Nakayama's lemma, the Statement 1 in the above link, we get $r\in 1+I$, such that $rN=0$, but $r\not \in I$ and $R$ is local implies $r$ is a unit, so $N=0$.
Remarks. 1) To get the choice of $k$, we can first assume $k=\dim_{R/I}(M/IM)$, then use the Statement 4 in the above link to lift the basis of $M/IM$ to get a minimal set of generators of $M$.
2) A deep theorem of Kaplansky says that any projective modules (not necessarily finitely generated) over a local ring is free.
Best Answer
To completely understand the +1 answer of Martin Brandenburg, I had to add a few details for myself. I decided to document the result in this answer:
Let a finite generating system of $P$ be given by $v_1,\ldots,v_n$. Set $F = R^n$ and $\varphi : F \to P$, $(x_1,\ldots x_n) \mapsto \sum_{i = 1}^n x_i v_i$. Since $P$ is projective, the epimorphism $\varphi$ splits, meaning that there is a monomorphism $\psi : P \to F$ such that $$\varphi\circ\psi = \operatorname{id}_P.$$
Now we check that $$F = \operatorname{im}\psi \oplus \ker\varphi.$$
Application of the homomorphism theorem to the monomorphism $\psi$ yields $P\cong\operatorname{im}\psi$, so $$P \oplus \ker\varphi \cong F = R^n$$ is finitely generated and free.