$$\require{AMScd}$$ $$\newcommand{\CC}{\mathbb{C}} \newcommand{\RR}{\mathbb{R}} \newcommand{\Hdr}{H_{\mathrm{dRh}}} \newcommand{\tensor}{\otimes} \newcommand{\Ohol}{\mathcal{O}}$$
Please excuse that this rather long question is essentially of the type "proof verification" which, I know, is frowned upon at this site. But I have no possibility to ask anyone in this case personally and by studying literature from the internet could not come to a final conclusion by myself.
Let $f: X \to S$ be a smooth projective map of smooth varieties over $\CC$, with $S/k$ a one-dimensional smooth affine curve over $\CC$.
Because of the Hodge-Decomposition for each $s \in S$, closed point, we have
$$ \Hdr^r(X_s / s) = \bigoplus_{p+q = r} H^q(X_s, \Omega_{X_s|k(s)}^p) = \bigoplus_{p+q = r} H^q(X, \Omega_{X|S}^p \tensor k(s)) $$
Now all $X_s \cong X_{s'}$ are isomorphic as complex manifolds, because $f:X \to S$ is a proper submersion and so by the Ehresmann fibration theorem a fibration.
So the ranks $\mathrm{rank} (H^q(X_s, \Omega_{X_s|k(s)}^p) = H^q(X, \Omega_{X|S}^p \tensor k(s)))$ which are in principle only upper semicontinous (see Hartshorne's chapter on semicontinuity theorems) are constant for varying $s \in S$.
Therefore again by the semicontinuity theorems, we have that $R^q f_* \Omega_{X|S}^p$ is a vector bundle on $S$ and we have isomorphisms
$$ H^a(X_s, \Omega_{X_s|k(s)}^b) = (R^a f_* \Omega_{X|S}^b) \tensor_{\Ohol_S} k(s) $$
for all $s \in S$ and $a,b \geqslant 0$ ("Grauert's theorem").
Now it is $\Hdr^r(X/S) = \RR^r f_*(\Omega_{X|S}^\bullet)$ where at the left stands the (algebraic) de Rham cohomology of $X/S$. The associated spectral sequence of hypercohomology for $\RR f_*$ is the so called Hodge--de Rham sequence (or Frölicher sequence): (here in a relative form for $X/S$):
$$ E_1^{pq} = ( R^q f_* \Omega_{X|S}^p ) \Rightarrow \RR^{p+q} f_* (\Omega_{X|S}^\bullet) = \Hdr^{p+q}(X/S) $$
I want to prove that $E_r^{pq}$ degenerates at the level $r=1$, and I want to use only the facts that the spectral sequence
$$ E_1^{pq}(k(s)) = (H^q(X_s, \Omega_{X_s|k(s)})) $$
degenerates at $r=1$ for every $s \in S$ and the assertions about the $R^a f_* \Omega_{X|S}^b$ from above.
So consider the differential $\delta := d_1^{pq}:E_1^{pq} \to E_1^{p+1,q}$. It is a map $\delta: R^q f_* \Omega_{X|S}^p \to R^q f_* \Omega_{X|S}^{p+1}$.
We tensor with $k(s)$ for $s \in S$ and get, that $\delta \tensor k(s): H^q(X_s, \Omega_{X_s|s}^p) \to H^q(X_s, \Omega_{X_s|s}^{p+1})$ vanishes because of degeneration of $E_1^{pq}(k(s))$.
Considering $\delta$ as element of $(R^q f_* \Omega^p_{X|S})^\vee \tensor (R^q f_* \Omega_{X|S}^{p+1})$ we get $\delta = 0$ therefore $E_2^{pq} = E_1^{pq}$.
Now we have to prove that $d^{pq}_r:E^{pq}_r \to E^{p+r,q-r+1}$ are all zero. Again I want to show this by the same consideration for all $s \in S$, tensoring with $k(s)$.
By induction the $E_r^{pq}$ are equal to $E_1^{pq}$. So one takes $d_r^{pq} \tensor k(s) = d_r^{pq}(k(s))$ (is this true? see my argument below) where $d_r^{pq}(k(s)): H^q(X_s,\Omega_{X_s|s}^p) \to H^{q-r+1}(X_s, \Omega_{X_s|s}^{p+r})$ is the map from the Hodge-de-Rham S-S for $X_s$ from level $r$ which we called $E_r^{pq}(k(s))$ further above.
So again we have $d_r^{pq}\tensor k(s) = 0$ and with the argument from above $d_r^{pq} = 0$.
QUESTION 1: Is this argument valid up to now?
To justify $d_r^{pq} \tensor k(s) = d_r^{pq}(k(s))$ I argue as follows:
One starts with the zero level page $E_0^{pq}$ from Cech-cohomology ($U_i$ an affine cover of $X/S$ induced by $(X \subseteq \mathbb{P}^N_S)/S$).
$$ E_0^{pq} = \prod_{i_0 <\cdots < i_q} \Gamma(U_{i_0\cdots i_q}, \Omega_{X|S}^p) $$
and writes the page $E_r^{pq}$ depending on it and on $d_0^{pq}$ as $E_r(E_0)$.
One has the diagram
\begin{CD} E_r(E_0(k(s))) @>{d_r(k(s)) = 0}>> E_r(E_0(k(s))) \\ @A=AA @A=AA \\ E_r(E_0 \tensor k(s)) @>{d_r(E_0 \tensor k(s))}>> E_r(E_0 \tensor k(s))\\ @A{\alpha_r}AA @A{\alpha_r}AA \\ E_r(E_0) \tensor k(s) @>{d_r \tensor k(s)}>> E_r(E_0) \tensor k(s) \end{CD}
In this diagram the $\alpha_r$ maps a "Zig-Zag" in $E_0$, which represents an element of $E_r(E_0)$, tensored with $k(s)$, to a "Zig-Zag" in $E_0 \tensor k(s)$, belonging to $E_r(E_0 \tensor k(s))$. (The side conditions on the Zig-Zag in $E_r(E_0)$ are retained upon tensoring with $k(s)$).
Additionally we have diagrams (with $Z_r = \ker d_r$)
\begin{CD} E_r(E_0 \tensor k(s)) @<\supseteq<< Z_r(E_0 \tensor k(s)) @>>> E_{r+1}(E_0 \tensor k(s)) \\ @A\alpha_rAA @A\alpha'_rAA @A\alpha_{r+1}AA \\ E_r(E_0) \tensor k(s) @<{\supseteq}<< Z_r(E_0) \tensor k(s) @>>> E_{r+1}(E_0) \tensor k(s) \end{CD}
Now $\alpha_1$ is an isomorphism, and so $d_1 \tensor k(s) = d_1(k(s)) = 0$. By the argument further above this gives $d_1 = 0$ and $E_2 = E_1$. So $Z_1(E_0) = E_1(E_0)$ and in the last diagram above all horizontal morphisms are isomorphisms and $\alpha_1$ as much as $\alpha_1' = \alpha_1$ are isomorphisms too. So we conclude $\alpha_2$ is an isomorphism. It follows $d_2 \tensor k(s) = d_2(k(s))$ and so we proceed inductively.
QUESTION 2: Is this argument valid?
Assuming, that $E_r^{pq}$ really degenerates at $r=1$, we have a filtration of the abutment $F^p \Hdr^r(X/S)$ with short exact sequences
$$ 0 \to F^{p+1} \Hdr^r(X/S) \to F^p \Hdr^r(X/S) \to R^q f_* \Omega^p_{X|S} \to 0 $$
and therefore an isomorphism
$$ (*) \quad\quad \Hdr^r(X/S) = \bigoplus_{p+q = r} R^q f_* \Omega_{X|S}^p $$
because for an exact sequence of vector bundles $0 \to F' \to F \to F'' \to 0$ on $S$ always $F = F' \oplus F''$ holds (consider $\mathrm{Ext}^1(F'', F') = 0$).
QUESTION 3: Is this argument valid, and do we really have the isomorphism $(*)$?
