No. Take $X$ to be the union of two smooth, genus $0$ curves, say $X_1$ and $X_2$, that intersect in two nodes, say $p$ and $q$. Then $\text{Pic}(X)$ is an extension of $\mathbb{Z}^2$ by $\mathbb{G}_m$. So let $Y$ be a geometric $\mathbb{A}^1$-bundle over $X$ whose restriction to each of $X_1$ and $X_2$ is trivial, yet that is nontrivial on $X$. Identify $X$ with the zero section of this $\mathbb{A}^1$-bundle. The normal cone is a normal bundle, essentially just the data of $Y$ once again. Yet you cannot recover $Y$ from its restriction to the two irreducible components.
Edit. Vivek clarified that he would also like to keep track of the restriction maps between the cones on different strata. That would certainly make a difference in the example above. However, if one allows non-transversal intersections, as Vivek explicitly allows, then there are still counterexamples. For instance, consider a union of three concurrent lines in $\mathbb{P}^2$. The irreducible components are each $\mathbb{P}^1$, and the intersections between components are all one (reduced) point. Now there is simply no way to encode all invertible sheaves by the restriction on strata and restriction homomorphisms between them. (Of course if one allows nilpotent thickenings of the strata, one could encode these invertible sheaves.)