Here is the solution I came up with for a restricted version of the question, which is sufficient for my main purpose. I will not accept my own answer, because I am still interested in a reference, a more complete solution, or a simpler proof.
I will prove the following.
Lemma. -
Let $\mu$ be a measure on $\mathbb{S}^1$ which is singular with respect
to the Lebesgue measure and without atoms;
then for all $v\in L^2(\mu)$, and all $\beta<1$ there
is a sequence
of smooth functions $\varphi_n:\mathbb{S}^1\to \mathbb{R}$ such
that $\varphi_n' \to v$ in $L^2(\mu)$ and $\varphi_n \to 0$ in
$\mathrm{Hol}_\beta(\mathbb{S}^1)$.
Proof.
We first claim that when $I\subset [0,1]$ is an interval of
length $\ell$,
$w:I\to \mathbb{R}$ is measurable and $\mu$-essentially bounded by some
number $M$, and $\varepsilon>0$, there is a smooth function
$\varphi:I\to\mathbb{R}$ such that $\varphi$ and all its derivatives
vanish at the endpoints of $I$,
$\lVert \varphi \rVert_\alpha \le 2\ell^{1-\beta} M$ and
$\int_I (\varphi'-w)^2 \,d\mu \le \varepsilon^2\ell^2$
Let $\eta>0$ be arbitrary, to be chosen later on.
Since $\mu$ is concentrated on a $\lambda$-negligible set,
there is a finite set of intervals $I_1,\dots,I_k\subset I$
with disjoint interiors whose total length is less than $\eta$ and whose
complement
$J = I\setminus (I_1\cup\dots\cup I_k)$
is given by $\mu$ a mass less than $\eta$. Let $\bar w$ be
the function which:
is constant on each $I_i$, with value the $\mu$-average
of $w$ on $I_i$,
is constant on $J$, with value such that
$\int_I \bar w \,d\lambda =0$.
By taking $\eta$ small enough and by dividing the intervals $I_i$
into smaller intervals, we can ensure that $\int (w-\bar w)^2 \,d\mu$
is arbitrarily small.
Define a Lipschitz function $\bar\varphi$ by
$$\bar \varphi(x) := \int_a^x \bar w \,d\lambda$$
where $a=\min I$ is the starting point of $I$.
Let now $\varphi$ be a smooth approximation of $\bar\varphi$
in uniform norm, such that $\varphi'$ stays close to $\bar w$
in $L^2(\mu,I)$ norm (one might need to make $\varphi''$ very large,
but this has no incidence on the claim). Since $\mu$ has no atom,
we can further flatten down $\varphi$
at the endpoints of $I$ in order to have its derivatives
vanish while keeping $\varphi'$ close to $\bar w$ in $L^2$ norm.
This approximation can easily be performed in a way further ensuring
that $|\varphi'|$ is bounded by $M$.
The uniform norm of $\varphi$ is then bounded by $M\eta$,
and can thus be made arbitrarily small.
Moreover for all $x,y\in I$:
$$\frac{|\varphi(x)-\varphi(y)|}{|x-y|^\beta}
\le \lVert\varphi'\rVert |x-y|^{1-\beta}
\le M \ell^{1-\beta}$$
so that taking $\eta$ small enough, we get the desired control
$\lVert \varphi \rVert_\alpha \le 2\ell^{1-\beta} M$, and
the claim is proved.
Now, given $v$ and $\varepsilon>0$, choose $\bar v$ a
$\mu$-essentially bounded function that is $\varepsilon$-close
to $v$ in $L^2(\mu)$, call $M$ its essential bound,
then choose $\ell$ small enough to
ensure that $2\ell^{1-\beta} M<\varepsilon$. Divide $\mathbb{S}^1$ into
intervals of length $\ell$ and apply the claim to each of them.
The boundary conditions enable us to glue the smooth functions
defined on each interval into a smooth function $\varphi$
defined on $\mathbb{S}^1$, and we get that $\varphi'$
is $\varepsilon$-close to $\bar v$ in $L^2(\mu)$. Moreover
the variations of $\varphi$ on $\mathbb{S}^1$ are not greater than
its variations on each interval, so that
$\lVert \varphi\rVert_\beta <\varepsilon$. This proves the Lemma.
