For local noetherian rings, the henselization has much more direct algebro-geometric meaning than the completion since it is built from local-etale algebras. Hence, your statement that the completion provides more intuition than the henselization is just a matter of having less experience with henselizations. See Example 2 below for why there is no need in practice for anything like a "formal topology", which likely doesn't exist anyway. In particular, the answer #1 is likely "no", so #2 is then moot.
But the real reason this is all somewhat of a wild goose chase is that you are correct about #4: for excellent local noetherian rings, the deep Artin-Popescu approximation is the ultimate answer to nearly all questions about passing between their completions and their henselizations. This theorem implies that for any such ring $R$ and any finite system of polynomial equations in several variables over $R$, any solution in the completion can be approximated arbitrarily closely (for the max-adic topology) by a solution in $R^{\rm{h}}$. (The real content, upon varying the system of equations, is that there is even a single solution in $R^{\rm{h}}$, let alone one that is "close" to the given solution in the completion.) However, that raw statement about systems of equations does not do justice to the significance, so let's illustrate with a couple of applications; see section 3 of Artin's IHES paper (http://archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1969__36_/PMIHES_1969__36__23_0/PMIHES_1969__36__23_0.pdf) for many more striking applications.
Example 1. A pretty algebraic application of this approximation theorem is that if $R$ is an excellent local normal noetherian domain then $R^{\rm{h}}$ is the "algebraic closure" of $R$ in the normal local noetherian completion $\widehat{R}$. Even for $R$ the local ring at the origin of an affine space over a field, I am not aware of a proof of this application without invoking Artin approximation.
Example 2. A geometric reason that Artin-Popescu approximation is so relevant is that it implies the following extremely useful fact: if $R$ is an exellent local noetherian ring and $A, B$ are two local $R$-algebras (with local structure map) essentially of finite type such that there exists an isomorphism $f:\widehat{A} \simeq \widehat{B}$ as $\widehat{R}$-algebras then $A$ and $B$ admit a common residually trivial local-etale neighborhood.
(In particular, $A^{\rm{h}} \simeq B^{\rm{h}}$ over $R$; we definitely do not claim that such an isomorphism of henselizations can be found which induces $f$ on completions.)
Even for $R$ a field, this is not at all obvious but is super-useful; e.g., it comes up when rigorously justifying the sufficiency of proving some facts about the geometry of specific singularities by studying analogous special cases with completions, such as in deJong's paper on alternations or various papers on semistable curves, etc.
It remains to address #3, under reasonable hypotheses on $R$. We will see that the first question in #3 is quite easy, and the second generally has a negative answer when $\dim R > 1$ (whereas the case $\dim R = 1$ is affirmative under reasonable assumptions on $R$, a fact used all the time in number theory; more on this below).
Now we come to the hypotheses that should have been made on $R$ for posing any questions of the sort being asked.
Presumably you meant for $R$ to be noetherian (max-adic completion would not be appropriate otherwise), and also excellent (or else the completion could fail to be reduced when $R$ is reduced, etc.). Moreover, let's forget about strict henselization and only consider henselization, since the latter has the same completion and is the correct "algebraic" substitute for the completion. And since henselization of an excellent local ring is excellent (see the Remark after Prop. 1.21 in the book "Etale Cohomology and the Weil Conjectures" for a proof, which was overlooked in EGA even though preservation under strict henselization is done in EGA), we may as well assume $R$ is henselian.
Since the category of finite etale algebras over a henselian local ring is naturally equivalent to that over its residue field, the invariance of etale fundamental group under passage to the completion for any henselian local noetherian ring is obvious. That is, the first question in #3 has an easy affirmative answer.
I do not know what is the precise intended meaning of the 2nd to last paragraph in the posting (since equicharacteristic 0 is a poor guide to the rich arithmetic structure of mixed characteristic $(0,p)$), but one might contemplate trying to use the Artin-Popescu approximation theorem to affirmatively settle:
Question. Let $R$ be an excellent henselian local domain that is normal, with fraction field $K$. Let $\widehat{K}$ denote the fraction field of the completion $\widehat{R}$. Do the Galois groups of $K$ and $\widehat{K}$ naturally coincide?
Despite whatever intuition may be derived from Artin-Popescu approximation, the possibly surprising answer to the Question is generally negative when $\dim R > 1$, even when $R$ is regular. (I only realized that after failing to prove an affirmative answer beyond dimension 1 in this way.)
Before discussing counterexamples, a few Remarks will be helpful to put things in context:
Remark 1: That the local noetherian ring $\widehat{R}$ is actually a normal domain requires a proof. One way to see it is to use Serre's homological characterization of normality and the fact that the flat map ${\rm{Spec}}(\widehat{R}) \rightarrow {\rm{Spec}}(R)$ has regular fibers (since $R$ is excellent); see the Corollary to 23.9 in Matsumura's "Commutative Ring Theory". Note also that when $\dim R > 1$, there is no "topological field" structure on $K$ and so $\widehat{K}$ is not a "completion" for $K$ in some intrinsic sense not referring to $R$; it is just notation for the fraction field of the completion of $R$.
Remark 2. In case $\dim R = 1$ the Question has an affirmative answer (even valid for rank-1 henselian valuation rings and valuation-theoretic completions thereof, without noetherian hypotheses; see 2.4.1--2.4.3 in Berkovich's paper in Publ. IHES 78). This does not require the approximation theorem, and is an instructive exercise in the use of Krasner's Lemma. The main task when $\dim R > 1$ is to try to replace Krasner's Lemma with an argument using the approximation theorem, bypassing the fact that $K$ (let alone its finite separable extensions) doesn't have a natural topological field structure when $\dim R > 1$; all such attempts are doomed to fail, as we'll see below.
Remark 3. Equality of the Galois groups (through the evident natural map) would imply via Galois theory that every finite separable extension of $\widehat{K}$ arises by "completing" a finite separable extension of $K$ and more generally that $E \rightsquigarrow E \otimes_K \widehat{K}$ is an equivalence of categories of finite etale algebraic over $K$ and $\widehat{K}$ respectively. This equivalence is very useful in number theory for the case $\dim R = 1$ when the answer is affirmative.
To see what goes wrong beyond dimension 1, consider a general $R$ as above, allowing $\dim R$ arbitrary for now and not yet assuming regularity. Let $K'/K$ be a finite separable extension, so by normality of $R$ the integral closure $R'$ of $R$ in $K'$ is module-finite over $R$. Since $R$ is henselian and $R'$ is a domain, it follows that $R'$ must be local (and henselian and excellent). Thus, $R' \otimes_R \widehat{R} = \widehat{R'}$ is a normal domain, and its fraction field $\widehat{K'}$ is clearly finite separable over $\widehat{K}$ with degree $[K':K]$.
In case $K'/K$ is Galois with Galois group $G$, it follows that $G \subset {\rm{Aut}}(\widehat{K'}/\widehat{K})$ with $\#G = [\widehat{K'}:\widehat{K}]$, so $\widehat{K'}/\widehat{K}$ is Galois with Galois group $G$. In this way, we see (by exhausting a separable algebraic extension by finite separable subextensions) that $K_s \otimes_K \widehat{K}$ is a Galois extension of $\widehat{K}$ with Galois group ${\rm{Gal}}(K_s/K)$. The Question is exactly asking if this is a separable closure of $\widehat{K}$.
By Galois theory over $\widehat{K}$, it is the same to ask that any given finite Galois extension $E/\widehat{K}$ is the "completion" (in the above sense based on integral closures of $R$) of some finite separable extension $K'/K$, necessarily of the same degree.
Now let's see that already for quadratic extensions one has counterexamples with $R$ any countable henselian excellent regular local ring with $\dim R > 1$ and ${\rm{char}}(K) \ne 2$, such as the henselization at any closed point on a smooth connected scheme of dimension $>1$ over a field not of characteristic 2; e.g., $R = \mathbf{Q}[x_1,\dots,x_n]_{(x_1,\dots,x_n)}^{\rm{h}}$ is a counterexample for any $n > 1$.
It is the same to show for such $R$ that the natural map $K^{\times}/(K^{\times})^2 \rightarrow \widehat{K}^{\times}/(\widehat{K}^{\times})^2$ is not surjective. Since $K^{\times}$ is countable, it suffices to show that the target is uncountable. This uncountability feels "obvious" when $\dim R > 1$, but we want to give a rigorous proof (e.g., to understand exactly how the condition $\dim R > 1$ and the completeness of $\widehat{R}$ are used).
The regularity of $R$ implies that of $\widehat{R}$, so $\widehat{R}$ is a UFD. Thus, the height-1 primes of $\widehat{R}$ are principal, corresponding to irreducibles in $\widehat{R}$ up to unit multiple. Any two irreducibles that are not $\widehat{R}^{\times}$-multiples of each other give rise to distinct elements in $\widehat{K}^{\times}/(\widehat{K}^{\times})^2$, by the UFD property. Thus, it suffices to show that $\widehat{R}$ contains uncountably many height-1 primes.
Complete local noetherian rings satisfy "countable prime avoidance": if an ideal $I$ is contained in a union of countably many prime ideals then it is contained in one of those primes. (This fact is originally a lemma of L. Burch. It can also be deduced easily from the Baire category theorem by using that such rings with their max-adic topology are complete metric spaces in which all ideals are closed, as noted in section 2 of the paper "Baire's category theorem and prime avoidance in complete local rings" by Sharp and Vamos in Arch. Math. 44 (1985), pp. 243-248.) By the UFD property, every non-unit in $\widehat{R}$ is divisible by an irreducible element and so lies in a height-1 prime. Thus, the maximal ideal $m$ of $\widehat{R}$ is contained in the union of all height-1 prime ideals. But $m$ is not contained in (equivalently, not equal to!) any of those height-1 primes since $\dim \widehat{R} = \dim R > 1$ (!), so by countable prime avoidance in $\widehat{R}$ it follows that $\widehat{R}$ must have uncountably many height-1 primes, as desired. (Related reasoning comes up in Example 2.4 in the paper of Sharp and Vamos mentioned above.)
Remark: For any $R$ as in the above counterexamples, and $B'$ the module-finite integral closure of $\widehat{R}$ in a quadratic extension of $\widehat{K}$ not arising from a quadratic extension of $K$, there is no finite $R$-algebra $B$ satisfying $B \otimes_R \widehat{R} \simeq B'$. This may seem to contradict Theorem 3.11 in Artin's IHES paper linked above, but it does not (since there is no reason that the proper closed non-etale locus in the base for the generically etale ${\rm{Spec}}(B') \rightarrow {\rm{Spec}}(\widehat{R})$ is contained in a proper closed subset of ${\rm{Spec}}(\widehat{R})$ that is the preimage of one in ${\rm{Spec}}(R)$).
Best Answer
You probably meant to assume $R$ and $S$ are noetherian. The answer is "no" to the initial hypergeneral part of the question. EDIT: In the 2nd half (below the long line), I now give a proof of an affirmative answer to the added part involving maps of affine spaces.
Counterexamples to the initial hypergeneral part can be made using 2-dimensional regular excellent local rings built from henselization and completion of local rings at $k$-points on smooth schemes over any field $k$.
Let $R$ be a noetherian henselian local ring, and $S = \widehat{R}$, so $\widehat{S}=S$. Let $\pi:{\rm{Spec}}(\widehat{R}) \rightarrow {\rm{Spec}}(R)$ be the natural map. Then in this case an affirmative answer to your question says exactly that $\pi_{\ast}$ is exact and $\pi^{\ast} \circ \pi_{\ast} \rightarrow {\rm{id}}$ is an isomorphism of functors (all on the category of constructible abelian etale sheaves on ${\rm{Spec}}(\widehat{R})$). In particular, every such sheaf on ${\rm{Spec}}(\widehat{R})$ would be the $\pi$-pullback of one on ${\rm{Spec}}(R)$. Clearly this cannot be true in general, so it is just a game to find a suitable counterexample to that.
Let $Z' \subset {\rm{Spec}}(\widehat{R})$ is a Zariski-closed set and $j':U' \hookrightarrow {\rm{Spec}}(\widehat{R})$ the open subscheme complementary to $Z'$. Consider the extension-by-zero $F' := j'_{!}(\mathbf{Z}/(n))$ for an integer $n > 1$. Suppose $F' = \pi^{\ast}(F)$ for an abelian etale sheaf $F$ on ${\rm{Spec}}(R)$.
Let $Z$ be the set of points $x \in {\rm{Spec}}(R)$ such that $F_x = 0$. This satisfies $Z' = \pi^{-1}(Z)$, so $Z$ is closed since $\pi$ is topologically a quotient map (as it is fpqc). Hence, if there is a closed set $Z'$ that is not the preimage of a closed set in ${\rm{Spec}}(R)$ then we have a counterexample.
Suppose $R$ is excellent, so the fpqc map $\pi$ is a "regular morphism" (flat and its fiber algebras that are regular and remain so after finite extension of the base field). Thus, if $Z$ is a reduced closed set in ${\rm{Spec}}(R)$ then the scheme-theoretic preimage $\pi^{-1}(Z)$ is reduced. Consider any radical ideal $J'$ in $\widehat{R}$ and let $Z' := {\rm{Spec}}(\widehat{R}/J')$. If $Z' = \pi^{-1}(Z)$ topologically for a closed set $Z$ in ${\rm{Spec}}(R)$ then by using the reduced scheme structure on $Z$ we would have $\pi^{-1}(Z) = Z'$ as schemes. In other words, for the radical ideal $J$ in $R$ corresponding to such a $Z$ we would necessarily have $J' = J\widehat{R}$.
Note that $J \otimes_R \widehat{R} \rightarrow J\widehat{R}$ is an isomorphism, so if $J'$ is invertible then necessarily $J$ is invertible. In particular, if $J' = r'\widehat{R}$ for $r'$ that is not a zero-divisor then necessarily $J = rR$ for some $r \in R$ that is not a zero-divisor. In such a situation, $r'$ would have to be a unit multiple of $r$ in $\widehat{R}$. Thus, if we can find $r' \in \widehat{R}$ that isn't a zero-divisor such that no unit multiple of $r'$ lies in the subring $R$ then we have our counterexample (using $Z' = {\rm{Spec}}(\widehat{R}/(r'))$ and the associated $F'$ as above).
In the special case that $R$ is regular, so $\widehat{R}$ is also regular, we're just looking for a nonzero $r' \in \widehat{R}$ having no unit multiple in $R$. Obviously in the dvr case this cannot be arranged, so we go on to dimension 2.
Take $R = k[x,y]_{(x,y)}^{\rm{h}}$ for a field $k$, so $\widehat{R} = k[\![x,y]\!]$. The ring $R$ is excellent, since passage to henselization preserves excellence (18.7.6, EGA IV$_4$). Since $k(\!(x)\!)$ is not algebraic over $k(x)$ (lazy way is to use $e^x$ and assume characteristic is 0), we can find $h \in x k[\![x]\!]$ that is not algebraic over $k(x)$. For such $h$, the irreducible element $r' = y - h(x)$ in $\widehat{R}$ will do the job.
Indeed, suppose that $r'$ admits a unit multiple $r \in R$. We will show that $h$ must be algebraic over $k(x)$. Clearly $R/(r) \rightarrow \widehat{R}/(r') = k[\![x]\!]$ is a completion map, so $R/(r)$ is a dvr. Since henselization is compatible with quotients, it follows by the link between transcendence degree and dimension for finitely generated domains over $k$ that $R/(r)$ is a direct limit of a directed system of local-etale extensions of local rings at $k$-points on regular curves over $k$, and these curves must all be quasi-finite over the affine $x$-line over $k$ (since $x \in k[\![x]\!]$ is transcendental over $k$). Thus, everything in $R/(r)$ is algebraic over $k(x)$, including the class of $y$. But mapping that into $k[\![x]\!]$ with $y \mapsto h(x)$ in there, we conclude that $h$ is algebraic over $k(x)$.
Now let's give an affirmative proof under some additional "smoothness" hypotheses as follows. The key point is to use Artin-Popescu approximation and to work at the level of derived categories. Also, one has to be extremely careful because certain schemes will intervene that (as far as I know) might fail to be noetherian or excellent, even though our original setup will involve only excellent schemes and smooth maps. (Maybe for some reasons which escape me, the non-noetherian concerns in the argument below cannot really happen?)
Setup: Let $\mathcal{X} \rightarrow \mathcal{Y}$ be a smooth map of noetherian schemes with $\mathcal{Y}$ excellent. Choose $y \in Y$ and $x \in X_y$ a $k(y)$-rational point. Let $X = {\rm{Spec}}(O_{\mathcal{X},x}^{\rm{h}})$, $Y = {\rm{Spec}}(O_{\mathcal{Y},y}^{\rm{h}})$. Let $X'$ and $Y'$ denote Spec of the corresponding completed local rings, and let $\pi:X \rightarrow Y$, and $\pi':X' \rightarrow Y'$ be the natural maps.
Claim: Let $F$ be a torsion etale abelian sheaf on $X$ whose torsion-orders are invertible on $Y$. Let $G \rightsquigarrow G'$ be shorthand for pullback on derived categories (of abelian etale sheaves) from $Y$ to $Y'$ or from $X$ to $X'$. Then the natural base change morphism $$R\pi_{\ast}(F)' \rightarrow R\pi'_{\ast}(F')$$ is an isomorphism for any torsion $F$ on $X$ whose torsion-orders are invertible on $Y$.
Proof: By Artin-Popescu approximation, since $\mathcal{Y}$ is excellent we know that the map $Y' \rightarrow Y$ is a limit of smooth maps. Hence, if we let $T = X \times_Y Y'$ and $p:T \rightarrow Y'$ be the projection then by the smooth base change theorem and standard limit stuff with etale sheaf theory, the natural map $R\pi_{\ast}(F)' \rightarrow Rp_{\ast}(G)$ is an isomorphism, where $G$ is the pullback of $F$ along the first projection.
I don't know if $T$ is noetherian, by the way, so I am tacitly using the good behavior of the limit formalism in etale sheaf theory without noetherian hypotheses (but even the proof of the smooth base change theorem in SGA4.5 leaves the noetherian framework due to such kind of fiber products intervening, so I suppose you don't mind).
The $y$-fiber of $p$ is identified with $X_y$, so naturally $x \in T$, and $O_{T,x}$ is a "partial henselization" of the local ring at a rational point on the special fiber of a finite type (even smooth) $Y'$-scheme. So even though I/we do not know if $T$ is noetherian, we do know that $O_{T,x}$ is noetherian, by the same reasoning which proves that passage to henselization preserves the noetherian property (namely, EGA 0$_{\rm{III}}$, 10.3.1.3).
Note that the henselization of $O_{T,x}$ coincides with that of a local ring $R$ on a finite type (even smooth) $Y'$-scheme, and $Y'$ is excellent (as for any complete local noetherian ring: IV$_2$, 7.8.3(iii)), so $R$ is excellent. Consequently $R^{\rm{h}}$ is excellent (IV$_4$, 18.7.6), so $O_{T,x}$ has excellent henselization. Now it is not true that excellence descends from the henselization (counterexample in IV$_4$, 18.7.7), but this is for geometric reasons related to failure of being universally catenary, and we don't actually care about that.
Indeed, what matters for the following is geometric regularity of formal fibers of $O_{T,x}$, which is to say that the flat map of noetherian schemes ${\rm{Spec}}(O_{T,x}^{\wedge}) \rightarrow {\rm{Spec}}(O_{T,x})$ is a regular morphism (since Artin-Popescu approximation is about regular morphisms being a limit of smooth morphisms, which in practice is easiest to remember under the banner of "excellence" but is not strictly a necessary condition). So all we care is to know that geometric regularity of formal fibers descends from the henselization, and that is fine; see IV$_4$, 18.7.4.
The upshot is that $O_{T,x}$ is noetherian and its completion morphism is regular.
Now comes the crux of the matter: I claim that the map $h:X' \rightarrow T$ (or rather, its factorization through ${\rm{Spec}}(O_{T,x})$) is computing the completion of the noetherian local ring $O_{T,x}$. (This is the step at which our argument breaks down in the setting of the counterexamples above! Indeed, in that setting the map in the role of $h$ would be akin to a graph morphism, super-far from an isomorphism.)
In view of the local structure theorem for smooth morphisms (applied to a smooth $Y'$-scheme whose "partial henselization" at a suitable point computes $T$), our hypothesis that $k(x) = k(y)$, and the fact that completion of a local noetherian ring is insensitive to "partial henselization", justifying our assertion about $h$ amounts to the following down-to-earth observation: if $A$ is a henselian (e.g., complete) local noetherian ring, $B = A\{x_1,\dots,x_N\}$ is the henselization at the origin of the special fiber of an affine space over $A$, and $B'$ is the local ring at the origin on the special fiber of the ring $\widehat{A} \otimes_A B$ (a ring I/we do not know to be noetherian, but the local ring $B'$ certainly is, as explained above), then the natural map $B' \rightarrow \widehat{A}[\![x_1,\dots,x_N]\!]$ is the completion of $B'$. This verification is a simple exercise using the compatibility of henselization and quotients.
From our description of $h$ via completion of a local noetherian ring having geometrically regular formal fibers, $h$ is a regular morphism. Thus, by Popescu's theorem, $h$ is a limit of smooth maps. Hence, by the acyclicity theorem for smooth maps and the usual limit games (which do not require knowing $T$ to be noetherian), it follows that the natural map $G\rightarrow Rh_{\ast}(h^{\ast}G)$ is an isomorphism. But $h^{\ast}(G)=F'$, so we have a natural isomorphism $$R\pi_{\ast}(F)' \simeq Rp_{\ast}(G)=Rp_{\ast}Rh_{\ast}(F')=R(p \circ h)_{\ast}(F').$$ But $p\circ h=\pi'$, and as such it is a standard exercise to check that the composite isomorphism we have made is the base change morphism of initial interest. QED