Here is a partial solution: modulo a problem of constructing "sufficiently generic" elements in the maximal ideal of a reduced noetherian local ring of dimension > 1 (in a sense made precise at the end in terms of associated primes and vanishing loci, and which might require some care when the residue field is finite), I claim it suffices to handle passage to residually trivial local-\'etale extensions (or what comes to the same thing, as we'll see below, henselization instead of completion). I'm not sure if this is any easier than completion, but it certainly can't be harder! I will use Artin-Popescu approximation (which wasn't known in 1980, the time of the Greco-Traverso paper). Let $R$ be a weakly normal reduced excellent local ring. Its henselization $R^{\rm{h}}$ not only inherits reducedness, but also inherits excellence (EGA IV$_4$, 18.7.6).
Let's reformulate the definition of weak normality in more convenient terms: the condition is that any finite radiciel surjection $X \rightarrow Y := {\rm{Spec}}(R)$ that is an isomorphism over a dense open is an isomorphism when $X$ is reduced. Now consider a finite radiciel surjection $X_1 \rightarrow Y_1 := {\rm{Spec}}(R^{\rm{h}})$ that is an isomorphism over a dense open, with $X_1$ reduced. Since $R^{\rm{h}}$ is a direct limit of residually trivial local-\'etale neighborhoods of $R$, by "spreading out" techniques from EGA IV$_3$, sections 8 & 9 there is a residually trivial local-\'etale cover $Y' \rightarrow Y$ and finite radiciel surjection $X' \rightarrow Y'$ that is an isomorphism over a dense open and induces $X_1 \rightarrow Y_1$ after base change, with $X'$ reduced. Thus, if all such $Y'$ were weakly normal then $X' \rightarrow Y'$ would be an isomorphism and so $X_1 \rightarrow Y_1$ would be too. In other words, if we knew weak normality were inherited by residually trivial local-\'etale extensions then the henselization would be weakly normal. I will now assume we knew such inheritance does hold, so we can replace $R$ with its henselization.
Suppose the completion $\widehat{R}$ (which we know is reduced, and of course excellent) were not weakly normal. Then there would be a finite radiciel surjection $T \rightarrow \widehat{Y} := {\rm{Spec}}(\widehat{R})$ that is an isomorphism over a dense open but not an isomorphism, with $T$ reduced. Thus, there is some infinitesimal fiber whose structure map to the base is not an isomorphism, say $n$th-order with $n \ge 1$. By A-P approximation for the excellent henselian $Y$, there is a local-smooth cover $Y' \rightarrow Y$, a $Y$-map $s:\widehat{Y} \rightarrow Y'$, and a finite radiciel surjection $T' \rightarrow Y'$ which is an isomorphism over a dense open $U' \subseteq Y'$ such that $T'$ is reduced and the pullback $T'_{\widehat{Y}}$ coincides with $T$ on $n$th-order fibers. Thus, if we pull back $T'$ along a $Y$-point $s'$ of $Y'$ which approximates $s$ to $n$th order (as we may certainly do, from the local structure theorem for smooth morphisms) then we arrive at a finite radiciel surjection $T'' \rightarrow Y$ which is an isomorphism over the $s'$-pullback of $U'$ and not an isomorphism on some infinitesimal special fiber. Since $Y$ is reduced and in fact weakly normal, to get a contradiction it suffices to arrange that ${s'}^{-1}(U')$ is not only dense in $Y$ (i.e., contains all generic points) but also that the corresponding open in $T''$ is schematically dense (so $T''$ inherits the reducedness from that open subscheme).
In other words, we are now faced with a problem having nothing to do with weak normality and everything to do with making a good choice for $s'$ to not lose contact with "dense open" downstairs and "reducedness" upstairs. Since $T' \rightarrow Y'$ is a finite radiciel surjection onto a reduced local ring with $T'$ also reduced, we can find a global function $f$ on $Y'$ that is non-vanishing at all generic points of $Y'$ and $T'$, hence nowhere a zero-divisor in both places. The issue is to find the "slice" $s'$ of the local-smooth (residually trivial) $Y$-scheme $Y'$ so that $s'$ is specified to $n$th order (from the $\widehat{Y}$-point) but $s'$-pullback preserve that property of $f$ being nowhere a zero-divisor both downstairs and upstairs.
If the base is a field there is nothing to do in the original problem, so we can assume the base $Y$ has positive dimension, so the local-smooth $Y'$ has dimension at least 2 (otherwise $Y' = Y$ and again there's nothing to do). In a reduced noetherian local ring with dimension > 1,
If we build up the slice $s'$ one step at a time (going through the "smooth coordinate" specializations into the base $Y$ approximating the given $\widehat{Y}$-point one at a time), standard arguments with associated primes and symmetry of ${\rm{Tor}}^1$ turn this into a task of prime avoidance: since we're allow to tweak things (for approx. purposes) beyond $n$th order in the closed point of the base, we are left having to solve the following general problem: if $A \hookrightarrow B$ is a module-finite radiciel extension of reduced noetherian local rings of dimension at least 2 and with the same total rings of fractions, we need to find $a \in A$ which is a non-zero-divisor in both $A$ and $B$ such that (i) its associated primes in both rings lie outside some specified nowhere-dense closed subset $Z$ of their "common" Spec and (ii) for any other given $a'$ in the maximal ideal of $A$ and chosen generating set $\{a_1,\dots,a_n\}$ of an ideal for $Z$ a "generic" linear combination of the $a_i$ using a fixed set of representatives for the residue field can be added to $a'$ to get an element non-vanishing at the generic points and the associated points for $a$ viewed in each of the two rings. (So you see, the case of finite residue field needs more care; but when you unravel where this is coming from, you may find a better way to express it to avoid that problem.)
