The answer is no, by a construction using the axiom of choice.

We shall build a counterexample set $A$ by a transfinite recursive
process of length continuum. At each stage, we shall promise that
certain elements are in $A$, in order to ensure that $A$ will be non-null, and that other elements are not in $A$, in such a way so as to prevent $\bigcup_n f_n(A)$ from containing a particular positive-measure Borel set. But
at each stage, we will have made fewer than continuum many such
promises altogether, and this fact will enable the construction to proceed to later stages.

To begin the construction, observe that there are continuum many Borel functions and therefore continuum many countable families $\{f_n\}$ of bijective
Borel functions that you consider, and also there are continuum many
Borel sets. So let us fix a well-ordered enumeration
$\langle
\{f_n^\alpha\}_n,B_\alpha\rangle$, for $\alpha<\mathfrak{c}$, of all pairs of such objects.

At stage $\alpha$, we consider first the possibility that $B_\alpha$ might be a
measure-zero Borel set containing the set $A$ we aim to construct. In order to prevent this, if $B_\alpha$ is measure
zero, then there must be some $c_\alpha\notin B_\alpha$ about which
we have not yet made any promises, and we promise now that
$c_\alpha\in A$. This will ensure that $A$ is not contained in this particular Borel measure-zero set, and therefore, since all such Borel measure-zero sets will eventually be considered, it will ensure that $A$ does not have measure zero.

Next, still at stage $\alpha$, we consider the possibility that $B_\alpha$ might be a positive-measure set contained in $\bigcup_n
f^\alpha_n(A)$. Since we have made so far fewer than continuum many
promises about $A$, it follows that we have promised fewer than
continuum many elements altogether to be in this union. But if $B_\alpha$ has
positive measure, then it must have size continuum, and so there is
a real $b_\alpha\in B_\alpha$ about whose pre-images
$a_\alpha=(f^\alpha_n)^{-1}(b_\alpha)$ we have not yet made any promises.
Let's now promise that none of these particular $a_\alpha$ are in
$A$, which is countably many additional promises at this stage.
This will ensure that $\bigcup_nf^\alpha_n(A)$ does not contain
this particular positive-measure Borel set.

By design, the construction ensures that $A$ is not measure zero,
yet there is no positive-measure Borel set contained in $\bigcup_n
f_n(A)$ for any Borel bijections $f_n$. And so $A$ has the features necessary to be the desired kind of counterexample.