(Joel's answer appeared as I was typing this.)
I think the answer is no.
Suppose to the contrary there exists a nonmeager set $A \subset \mathbb{R}$ which is the union of some chain $\{K_i\}_{i \in I}$ of nowhere dense sets. $A$ is separable, so we may enumerate a countable dense set $\{x_n\} \subset A$. Then we can find an increasing sequence $\{K_{i_n}\}$ with $x_n \in K_{i_n}$. Setting $K = \bigcup_n K_{i_n}$, by the Baire category theorem $K \ne A$, so there exists $x \in A \backslash K$. Now there must be some $K_j$ with $x \in K_j$. Now for each $n$ we certainly don't have $K_j \subset K_{i_n}$, so we must have $K_{i_n} \subset K_j$ since the $K_i$ are a chain. Thus $K \subset K_j$, but then $K_j$ contains all the $x_n$ and so is not nowhere dense.