Andy:

A good reference for your questions is "Consequences of adding Cohen reals" by J. Steprans, in "Set Theory of the reals", Judah ed., Bar-Ilan University, 1993, pp. 583-617. Another reference is the excellent book "Set theory: on the structure of the real line" by Bartoszynski and Judah.

Adding a Cohen real makes the ground model reals have measure zero. I mentioned this in another answer, and it is similar to what you write in your post. Briefly: Any nontrivial countable forcing notion is equivalent to Cohen. For example, fix $\epsilon>0$ and let ${\mathbb P}_\epsilon$ be the collection of subsets $A$ of ${\mathbb R}$ that are the finite union of open intervals with rational endpoints such that the measure of $A$ is less than $\epsilon$. Ordered by (reverse) inclusion, this poset is equivalent to Cohen forcing. The generic is an open set of measure $\epsilon$ that covers the ground model reals. Since $\epsilon$ was arbitrary, the result follows. Steprans's paper contains a proof of a strengthening of this fact.

On the other hand, the ground model reals are not meager in the extension. In fact, any second category set of reals in the ground model is still second category after adding a Cohen real. (This is also proved in Steprans's paper.)

Finally, it is a theorem of Velickovic and Woodin that if $V\subseteq W$ are models of set theory, and for every countable $X\in W$ with $X\subseteq{\mathbb R}^V$ there is a $Y\in V$ countable in $V$ with $X\subseteq Y$, then, if there is a perfect set consisting of reals from $V$, then every real is in $V$. In particular, the ground model reals cannot contain a perfect set after adding a Cohen real. For a reference, see Velickovic-Woodin, "Complexity of reals in inner models of set theory", Annals of pure and applied logic, 92 (1998), 283-295.

[**Edit**: Martin Goldstern has now posted a nice argument by a more natural line of thought than my naive approach.]

Here is a sketch of an argument I *think*, if one manages to flesh out, should show that adding a Cohen real makes ${\mathbb R}^V$ be strong measure zero. By the way, I was unable to find an explicit mention of this result or a proof in any of the references I mentioned (or elsewhere). Laver's original paper ("On the consistency of Borel's conjecture", Acta Math., 137 (1976), no. 3-4, 151–169) only mentions in passing (page 155) that adding a Cohen real makes ${\mathbb R}^V$ strong measure zero. There may not be an explicit reference in print, actually.

We need the following basic property of Cohen forcing:

If $c$ is Cohen over $V$ and $g:\omega\to\omega$ in $V[c]$, then there is an $h:\omega\to\omega$ in $V$ such that $g(n)\lt h(n)$ infinitely often.

Let $c$ be Cohen over $V$ and let $\vec\varepsilon=(\varepsilon_0,\varepsilon_1,\dots)$ be a sequence in $V[c]$ of positive reals. As you observe, if $\vec\varepsilon\in V$, then
$$ {\mathbb R}\cap V\subseteq \bigcup_n I^n_{c(n)} $$
where (for all $n$) the sequence $(I^n_m\mid m<\omega)$ lists (in $V$) all intervals with rational endpoints and length at most $\varepsilon_n$.

Suppose now that $\vec\varepsilon\in V[c]\setminus V$. Without loss, each $\varepsilon_m$ has the form $2^{-n_m}$ for some positive integer $n_m$, and the sequence is strictly decreasing. Fix $h\in V$ such that $A=\{m\mid n_m\le h(m)\}$ is infinite.

For each $n$, let $(I_{n,m}\mid m\in\omega)$ list (in $V$) all intervals with rational endpoints and length at most $2^{-h(m)}$. The argument "reduces" then to proving the following key fact:

A genericity argument should show that, in fact,
$$ {\mathbb R}\cap V\subseteq \bigcup_{n\in A}I^n_{c(n)}. $$

If we manage to prove this, since the sequence of numbers $n_m$ is increasing, we are done.

Note that another basic property of Cohen forcing gives us that there is a real $t$ in the ground model coding a Borel function $f$ such that $A=f(c)$. To make the argument work, something stronger (like $f$ recursive with the real $t$ as oracle) seems needed. I don't see how to make the result go through for arbitrary Borel $f$.