Stefan's answer pointed me in the right direction, and then talking it over with prof. Leo Harrington we've got an answer:
A complete Boolean algebra $\mathbb{B}$ never adds a dominating real iff for any collection $\{ u _{m,k} : m, k \in \omega \} \subset \mathbb{B}^+$ the following weaker form of weak $(\omega ,\omega )$-distributivity holds:
$\prod _{m \in \omega} \sum _{k \in \omega} u _{m,k} = \sum _{f \in \omega ^{\omega}} \prod _{n \in \omega} \sum _{n < m < \omega} \sum _{k < f(m)} u _{m,k}$
For the reverse implication, suppose "weak weak $(\omega ,\omega )$-distributivity" holds, and for contradiction let $u \neq 0$ be the Boolean value of the sentence "there exists a dominating real," and let $\dot{g}$ be a name witnessing this, i.e. the sentence "$\dot{g}$ is a dominating real" has Boolean value $u$. Define $u _{m,k} = || \dot{g} (m) = k ||$. Now if $G$ is any $\mathbb{B}$-generic filter containing $u$, noting that the left side of the distributivity identity is (at least) $u$, we know that the right side belongs to $G$. It's then not hard to see that:
$(\exists f \in (\omega ^{\omega})^V )(\forall n \in \omega )(\exists m > n)(\exists k < f(m))(u _{m,k} \in G)$
which is to say that there's a real $f$ in the ground model such that:
$(\forall n)(\exists m > n)(\dot{g}^G (m) < f(m))$
so $f$ is not dominated by $\dot{g}$, contradiction.
For the forward implication, it should suffice to show it in the case where for each $m$, the set $\{ u _{m,k} : k \in \omega \}$ is an antichain with least upper bound $u$ independent of $m$ (I haven't checked this detail personally). So let $\{ u _{m,k}\}$ be such a collection for which the identity fails. Consider the name:
$\dot{g} = \{ (u _{m,k}, (m,k)) : m, k \in \omega \}$
It's not hard to see that the right side of the identity is at most $u$, so assuming the identity fails it's strictly less than $u$, so since $\mathbb{B}$ is separative there's a generic $G$ containing $u$ avoiding the right side of the identity. It's not hard to see from here that $\dot{g}^G$ will dominate all the ground model's reals.
I should add that if we think of $u _{m,n}$ as saying "$\dot{g}(m) = n$" and replace the Boolean operations with the corresponding quantifiers, then the left side says "$\dot{g}$ is a real," and the right side says "$\dot{g}$ doesn't dominate every real in the ground model." This suggests how we can characterize forcings that don't add any unbounded reals, for example, namely the following identity holds:
$\prod _{m \in \omega} \sum _{k \in \omega} u _{m,k} = \sum _{f \in \omega ^{\omega}} \prod _{m \in \omega} \sum _{k < f(m)} u _{m,k}$
Forcings that don't add any reals are precisely those that satisfy the following identity:
$\prod _{m \in \omega} \sum _{k \in \omega} u _{m,k} = \sum _{f \in \omega ^{\omega}} \prod _{m \in \omega} u _{m,f(m)}$
You can easily generalize this to talking about functions $\kappa \to \lambda$; the above two results so generalized are precisely Theorem 15.38 and Lemma 15.39 in Jech, "Set Theory".
(An attempt at an answer, and also my first posting here. Thanks to Andres Caicedo for the reformatting.)
I claim that a single Cohen real makes the set of old reals strong measure zero.
Reals are functions from $\omega$ to 2.
Let ${\mathbb C}$ be Cohen forcing, and let $c$ be the name of the generic real.
Let $(n_k)$ be a sequence of ${\mathbb C}$-names for natural numbers.
I will find a sequence $(s_k)$ of names for finite $01$-sequences
($s_k$ of length $n_k$)
such that ${\mathbb C}$ forces: every old real is in some $[s_k]$.
Let $D_k$ be a dense open set deciding the value of $n_k$ and
containing only conditions of length at least $k$.
Say, each $q$ in $D_k$ decides that the value of $n_k$ is $f_k(q)$, where
$f_k$ is a function in the ground model defined on $D_k$.
Each $f_k$, and also the sequence $(f_k)$, is in $V$.
Now we work in the extension.
(The point is that even though we now know the actual values of
$n_k$, we will play stupid and use the names only, plus the minimal
amount of information that we need from the generic real.
This lets us gauge exactly how much information from the
generic we need.)
In the extension I will define a sequence $(i_k)$ of natural numbers.
Let $i_k$ be the minimal $i$ such that $c \mathord\upharpoonright i$ is in $D_k$,
where $c\mathord\upharpoonright i = c$ restricted to $i$.
(So $i_k$ is at least $k$.)
For each $k$ we now define a $01$-sequence $s_k$ of length $n_k$ as follows: Take $n_k$ successive bits from the Cohen real $c$, starting at position $i_k$. (Formally: $s_k(j) = c(i_k+j)$ for all $j\lt n_k$.)
I claim that "every old real is in some $[s_k]$" is forced.
Assume not, so let $p$ force that $x$ is not covered.
Let $k$ be larger than the length of $p$. So $p$ not in $D_k$.
Extend $p$ to $q$ so that $q$ is in $D_k$, $q$ minimal.
Let $l$ be the length of $q$.
So $q$ forces that $i_k$ is exactly $l$. Also $q$ forces that $n_k = f_k(q)$.
Now extend $q$ to $q'$, using the first $f_k(q)$ bits of $x$.
So $q'$ is stronger than $q$, and $q'$ forces that $s_k$ is an initial
segment of $x$.
mg*
Best Answer
I think that Groszek and Slaman's result (see https://www.jstor.org/stable/421023?seq=1) gives a satisfying answer to your question.
Groszek and Slaman's result says that given any inner model $M$ of $ZFC$, $\mathbb{R}\subset M$ if and only if there is a perfect set $A\subset M$.
A immediate conclusion of the result is that for any inner model $M$ of $ZFC$ and an $M$-Borel set $A$, $A$ remains to be Borel in any extension $N$ of $M$ with $\mathbb{R}^M\neq \mathbb{R}^N$ if and only if $A$ is countable in $N$.