Tldr; jump to the $(\ast)$.
The category $Top$ of topological spaces is complete and cocomplete. That is, all limits and colimits exist in the category. On the other hand, the homotopy category $hTop$, formed by quotienting out the (weak) equivalences, is neither complete nor cocomplete. In fact while it has products and coproducts, it has very few other limits or colimits. See here for some examples of pullback/pushout squares which do not exit in $hTop$.
The point is that the strict fibre of an arbitrary map $f:X\rightarrow Y$ over a point $y\in Y$ is the equaliser of the two maps $f,c_y:X\rightrightarrows Y$, where $c_y$ is the constant map at $y$. Now this is not a good homotopical notion, in general, due to the lack of limits in $hTop$. It is a theorem that if a category has equalizers and finite products then it has all finite limits, and we have already noted above that $hTop$ does not even have all pullbacks. Since it has products $hTop$ cannot have all equalisers.
To see exactly what goes wrong consider the fact that the strict fibres of $f:X\rightarrow Y$ over the different points of $Y$ will not in general have the same homotopy type (even keeping within the same path component). For instance consider the map $S^1\rightarrow \mathbb{R}$ induced by projecting onto the first coordinate; the fibres are either empty, have one point, or have two points.
A sufficient condition that all the strict fibres over all points in a given path component have the same homotopy type is that the map $f$ is a fibration. And this is what the homotopy fibre does: it replaces the map $f$ by a fibration $p_f:E(f)\rightarrow Y$ and then takes its strict fibre over a given point. For everything to make sense we require some comparison map $j_f:X\rightarrow E(f)$ which is a homotopy, or at least weak, equivalence, and that everything should be natural - in the mathematical sense - at least up to homotopy.
Note that this 'up-to-homotopy' comparison map $j_f$ is the best we can do, for if $j_f$ were a homeomorphism, then the map $f$ would already be a fibration, and its homotopy fibres would just be its strict fibres. Thus if it were always possible to replace a map by a fibration up to homeomorphism, then there would be no need for homotopy fibres to begin with.
$(\ast)$ Now, the point is that the homotopy fibre of a map $f:X\rightarrow Y$ does not really live in $Top$. Since at best we can ask for the map $j_f$ to be a homotopy equivalence, the homotopy fibre of $f$ really lives in the homotopy category $hTop$ where it makes sense to consider the objects $X$ and $E(f)$ as the same. And now in the homotopy category everything is perfectly canonical, since $X$ and $E(f)$ are the same object as seen through categorical eyes. For example in the path space fibration we no longer have to pick an evaluation, since all these maps are homotopy equivalent, and thus represent the same morphism in $hTop$.
Thus we realise that the object $E(f)$ is nothing but an auxiliary construct. It is a point set lift of the problem in $hTop$ to a problem in $Top$. It allows us to create a more understandable model for the problem by using spaces with points, and maps defined on elements, rather than having everything 'up-to-homotopy'. However there is no canonical way to achieve this. For, just as the strict fibres of an arbitrary map $f:X\rightarrow Y$ may be badly behaved, the fibres of the projection $Top\rightarrow hTop$ are not uniform. Asking for a canonical section of this map is too much.
Thus we content ourselves with mathematically natural (again, up to homotopy) constructions point-set constructions, happy with the knowledge that once we project the problem back to $hTop$ we truly do have something safe and canonical.
Let $\mathscr{C}$ be any $\infty$-category with finite colimits and a terminal object $*$. We denote $\mathscr{C}_*=*/\mathscr{C}$. Let $\mathsf{A}$ be the diagram shape $2\leftarrow 0\to 1$, and let $i_0\colon[0]\to\mathsf{A}, 0\mapsto 0$ be the inclusion of the initial vertex, with corresponding right Kan extension $\mathrm{Ran}_{i_0}\colon\mathscr{C}\to\mathrm{Fun}(\mathsf{A},\mathscr{C})$ (informally sending $X$ to a diagram $*\leftarrow X\to *$). Let $\mathsf{Sq}$ be the commutative square category
$$
\require{AMScd}
\begin{CD}
0 @>>> 1\\
@VVV @VVV\\
2@>>> 3
\end{CD}
$$
with corresponding inclusion functor $i\colon\mathsf{A}\to\mathsf{Sq}$. This gives us a corresponding left Kan extension $\mathrm{Lan}_{i}\colon\mathrm{Fun}(\mathsf{A},\mathscr{C})\to\mathrm{Fun}(\mathsf{Sq},\mathscr{C})$, which takes a colimit of the span to ''complete the square''. Let $\mathsf{E}$ be the one-arrow category $1\to 3$, with corresponding inclusion functor $j\colon\mathsf{E}\to\mathsf{Sq}$. We define a functor $\widetilde{\Sigma}_+\colon\mathscr{C}\to\mathscr{C}_*$ as the composite
$$
\mathscr{C}\xrightarrow{\mathrm{Ran}_{i_0}}\mathrm{Fun}(\mathsf{A},\mathscr{C})\xrightarrow{\mathrm{Lan}_{i}}\mathrm{Fun}(\mathsf{Sq},\mathscr{C})\xrightarrow{j^*}\mathrm{Fun}(\mathsf{E},\mathscr{C})\simeq\mathrm{Ar}(\mathscr{C}),
$$
which lands in the full subcategory of the arrow category $\mathrm{Ar}(\mathscr{C})$ on the arrows with $*$ as domain. Hence $\widetilde{\Sigma}_+$ factors through the inclusion $\mathscr{C}_*\to\mathrm{Ar}(\mathscr{C})$. Denote by $\Sigma_+\colon\mathscr{C}\to\mathscr{C}_*$ the resulting functor. Writing $U\colon\mathscr{C}_*\to\mathscr{C}$ for the forgetful functor, we of course have $\Sigma\colon\mathscr{C}\to\mathscr{C}$ to be naturally equivalent to $U\Sigma_+$. In a similar way, we can formally define the suspension functor $\mathrm{Sigma}_*\colon\mathscr{C}_*\to\mathscr{C}_*$. Given an object $(X,x)$ of $\mathscr{C}_*$, the object $\mathrm{Sigma}_*(X,x)$ is naturally equivalent to $\Sigma X$ equipped with the basepoint $*\xrightarrow{x}X\to\Sigma X$. However, since $X\to\Sigma X$ factors as $X\to *\to \Sigma X$, where $*$ is the terminal object at position $1$ in $\mathsf{A}$ and $\mathsf{Sq}$, we find that this basepoint is naturally equivalent to the basepoint that $\Sigma X$ receives from its lift $\Sigma_+X$. In fact, we find that $\Sigma_*(X,x)\simeq\Sigma_+U(X,x)$ naturally in $(X,x)$.
This means that the diagram
of $\infty$-categories and functors commutes. This in turn implies that
$$
\mathrm{colim}(\mathscr{C}\xrightarrow{\Sigma}\mathscr{C}\xrightarrow{\Sigma}\ldots)\simeq\mathrm{colim}(\mathscr{C}_*\xrightarrow{\Sigma_*}\mathscr{C}_*\xrightarrow{\Sigma_*}\ldots),
$$
so the two definitions agree. (Note that, for general $\mathscr{C}$, neither of these $\infty$-categories will model spectrum objects in $\mathscr{C}$.)
That leaves the question why we prefer to work with the pointed version. I can think of a couple of reasons: firstly, pointed categories are easier to work with than unpointed ones. Having a zero map can for instance make it easier to make certain constructions natural. Secondly, the suspension functor $\Sigma$ as defined above is generally not a left adjoint even if $\mathscr{C}$ has finite limits, while $\Sigma_*$ is in that case (from now on, just assume $\mathscr{C}$ has finite limits). $\Sigma_+$ is also not a left adjoint: you really need to choose both natural basepoints of $\Sigma X$ as such and consider the lift $\Sigma_{++}\colon\mathscr{C}\to\mathscr{C}_{*\sqcup*/}$ to get a right adjoint functor, namely the loop functor $\Omega$. But as said, $\Sigma_*$ is just immediately a left adjoint. This is particularly helpful if you don't work $\infty$-categorically: if you work with model categories, then your main way to model $\infty$-functors is by building Quillen adjunctions. In particular, you want a good supply of adjoint $1$-functors, so you prefer to work with the pointed version. But also $\infty$-categorically we prefer adjoint functors over non-adjoint ones. Thirdly, the equivalence
$$
\mathrm{colim}(\mathscr{C}\xrightarrow{\Sigma}\mathscr{C}\xrightarrow{\Sigma}\ldots)\xrightarrow{\simeq}\mathrm{colim}(\mathscr{C}_*\xrightarrow{\Sigma_*}\mathscr{C}_*\xrightarrow{\Sigma_*}\ldots),
$$
defined above requires a choice, namely the choice in the definition of $\Sigma_+$ to let one, and not the other natural candidate for the basepoint be the actual basepoint. You might object, and say that if we let $\Sigma_{-}$ denote the functor we would get by choosing the other basepoint, then the fact that both $\Sigma_+$ and $\Sigma_{-}$ induce on colimits inverses to the canonical equivalence
$$
\mathrm{colim}(\mathscr{C}\xrightarrow{\Sigma}\mathscr{C}\xrightarrow{\Sigma}\ldots)\xleftarrow{\simeq}\mathrm{colim}(\mathscr{C}_*\xrightarrow{\Sigma_*}\mathscr{C}_*\xrightarrow{\Sigma_*}\ldots),
$$
in the other direction (the one induced by $U$) would imply that both $\Sigma_+$ and $\Sigma_{-}$ induce up to equivalence the same map on colimits. And while that is true, it is not up to canonical equivalence. So there is a very slight non-canonical thing occurring because we had to choose whether the north or the south pole of our suspension was the base point. And while these can be connected by a path, the choice of path is not canonically determined up to higher homotopies. In the pointed setting, everything is completely canonical. There probably are more reasons (also of less philosophical and more practical nature), but I'll keep it at this for now.
Best Answer
It is hard to say what Hatcher really means, unfortunately he isn't always precise in his formulations.
As far as I know we need a pointed fibration to get the eaxct sequence in your question. However, if you want to work with free fibrations $p : E \to B$, you can easily prove that for each well-pointed $(X,x_0)$ you get an exact sequence $$\langle (X,x_0), (F,e_0)\rangle \to \langle (X,x_0), (E,e_0)\rangle \to \langle (X,x_0), (B,b_0)\rangle$$
I guess this is what Hatcher has in mind since he mainly works with CW-complexes in Chapter 4 (he does not say anything explicitly about $X$).