The answer is no. Assume that such a measure $\mu$ exists.

First, since every singleton in $[0,1]$ is closed with empty interior, $\mu(\{x\}) = 0$ for all $x \in [0,1]$. Write $B_{x,\epsilon}$ for the open ball around $x$ of radius $\epsilon$ with respect to the standard metric on $[0,1]$. By countable additivity, for all $x \in [0,1]$, $\mu(B_{x,2^{-n}}) \to 0$. If we take an enumeration of the rationals $(q_i)_{i \in \mathbb{N}}$, for each $i$ there exists an $n_i$ such that $\mu(B(q_i,2^{-n_i})) < 2^{-i}$. So $D_1 = \bigcup_{i=1}^\infty B(q_i,2^{-n_i})$ is a dense open set with $\mu(D_1) \leq 1$.

By re-doing the previous construction, picking $\mu(B(q_i,2^{-n_i})) < 2^{-(i+k)}$, we can define dense open sets $D_k$ with $\mu(D_k) \leq 2^{-k}$. Now, by countable additivity $N = \bigcap_{k=1}^\infty D_k$ has measure zero. The set $[0,1]\setminus N$ is a union of closed sets with empty interior, *i.e.* a meagre set, so $\mu([0,1] \setminus N) = 0$ as well, so $\mu([0,1]) = 0$.

I see that Nate Eldredge was a bit quicker than I was, so I'll add that it *is* possible to find a finitely-additive probability measure whose null sets are exactly the meagre sets -- this is most easily done using the isomorphism between the Baire property algebra modulo meagre sets and the algebra of regular open sets.