Well, for a general ASG, the symmetric group $S_{n}$ is almost simple for $n > 4,$ and there is usually no prime $p$ for which property 1 above holds. For example, if the $p$-subgroup $X$ fixes more than $4$ points, $C_{S}(X)$ will have an alternating group as a component. It is possible to work out the order of the largest Sylow subgroup of $S_{n}$ or $A_{n}.$ This is often, but not always, the Sylow $2$-subgroup. However, in $A_{9},$ for example, a Sylow $3$-subgroup has order $81$ and a Sylow $2$-subgroup has order $64.$

As for the ASGs of Lie type, the outer automorphism groups of their generalized Fitting subgroups are solvable. This means that for a $p$-subgroup $X$ of simple $S$ which is $F^{*}(A)$ for some ASG $A$, any component of $C_{A}(X)$ would already be contained in $C_{S}(X)$ as $C_{A}(X)/C_{S}(X)$ is solvable. Hence in Lie type characteristic $p$ ASGs groups, we still have $F^{*}(C_{A}(X) ) = F(C_{A}(X))$ for $p$-subgroups $X.$ It is unusual, but there may be examples where $O_{q}(C_{A}(X)) \neq 1$ for some prime $q \neq p,$
but they should be well understood and documented, eg in Lyons Solomon books.

As for the last question, the outer automorphism group of a non-Abelian simple group is very well understood, assuming the classification of finite simple groups,(see the Atlas of Finite Groups for example) so it is easy to compare the order of any Sylow subgroup of an almost simple group and a Sylow subgroup of its generalized Fitting subgroup, since the almost simple group only differs from its generalized Fitting subgroup by the inclusion of outer automorphisms of its generalized Fitting subgroup.

*Answer to comment by user44924* : In the discussion of $p$-constraint above, I was considering $X$ as a subgroup of $S.$ If $X$ is a subgroup of $A$ with $Y = X \cap S \neq 1,$ then $C_{A}(X) \leq C_{A}(Y)$ and nothing much changes. However, if $X \cap S = 1$ ( a relatively rare situation, since we are talking of a characteristic $p$ Lie type group admitting an outer automorphism of order $p,$ and for large primes $p,$ this can only come from a field automorphism) things can be different. For example, if $x$ is a field automorphism of order $p$ of $S = {\rm PSL}(n,p^{p}),$ then $C_{S}(x) = {\rm PSL}(n,p),$ so $C_{S}(x)$ does have a component when $p > 3$ for example.