No. On $\mathbb P^1=\mathbb P^1(\mathbb C)$ we have $\Gamma(\mathbb P^1,\mathcal O_{\mathbb P^1}(-1))=\Gamma(\mathbb P^1,\mathcal O_{\mathbb P^1}(-2)=0$, but $O_{\mathbb P^1}(-1)$ and $O_{\mathbb P^1}(-2)$ are not isomorphic.
However on an affine algebraic variety $X$, the answer is "yes". There is an amazing equivalence of categories between $\mathcal O(X)$-modules and the so-called quasi coherent sheaves on $X$. It is denoted $M\mapsto \tilde M.$
In particular if you have a vector bundle $E$ on $X$, you can recover it (or rather its locally free associated sheaf) from $M=\Gamma(X,E)$ by this equivalence.
And this remarkable result is not even very difficult! ( Hartshorne, Algebraic Geometry, II Corollary 5.5) . And it is valid on any affine scheme!
Another interpretation
I have interpreted $\mathcal O(E)$ as the vector space $ \Gamma(X,E)$ of global sections of the bundle $E$.
However Donu Arapura and Qfwfq consider that the notation designates the sheaf of sections of the vector bundle $E$, that is the sheaf $\mathcal E$ associating to the open subset $U\subset X$ the vector space $\mathcal E(U)=\Gamma(U,E)$.
In that case the answer is : yes, that sheaf determines the bundle.
Indeed there is a canonical way to obtain from the sheaf $\mathcal E=\mathcal O(E)$ a holomorphic vector bundle $ Vec(\mathcal E) $ isomorphic to $E$.
Its fiber at $x\in X$ is the the $\mathcal O_{X,x}/ \mathfrak m_x=\mathbb C$- vector space $Vec(\mathcal E) [x]=\mathcal E _x/\mathfrak m_x \mathcal E_x$.
And the complex structure on $Vec(\mathcal E)=\bigsqcup Vec(\mathcal E) [x]$, is obtained from bijections with $U\times \mathbb C^r$ for all $U$'s on which $\mathcal E|U$ is free, that is isomorphic to $\mathcal O^r_U$.
Of course, there are verifications to be made, which are as straightforward as they are boring and unpleasant to write down explicitly...
Edit This latter interpretation is the one chritian had in mind, as he just stated in a comment.