Here is one way of proving the conjecture is true in general, using the modern method of weak factorization systems.
A weak factorization system has at its core two classes of maps the left class and the right class and they satisfy lifting properties with respect to each other. Here the right class is the class of Hurewicz fibrations. The left class is the class of trivial Hurewicz cofibration (i.e. DR pair). These classes define each other. The left class consists of all maps with the left lifting property with respect to all maps in the right class; the right class consists of all maps with the right lifting property with respect to the left class. So trivial Hurewicz cofibrations have the left-lifting property with respect to the Hurewicz fibrations. They are exactly the Hurewicz cofibrations which are also homotopy equivalences.
We won't really need the general notion, just a few special cases. You already know some trivial Hurewicz cofibrations. For example for any space $B$, the inclusion
$$B \times \{0\} \to B \times [0,1]$$
has the left lifting property with respect to any Hurewicz fibration (by definition), hence this is the first example of a trivial Hurewicz cofibration.
Also the trivial Hurewicz cofibrations are closed under several operations: composition, taking retracts, and cobase change (aka pushouts). Using these we can form new examples of trivial Hurewicz cofibrations. Let $f: B' \to B$ be any map. Then the inclusion of $B$ in the mapping cylinder
$$B \to B' \times [0,1] \cup^{B'\times \{1\}} B$$
is an example (a pushout of our previous example). We can also consider the inclusion on the other side
$$B' \times \{0\} \to B' \times [0,1] \cup^{B'\times \{1\}} B$$
If the map $f$ is a homotopy equivalence then this too is a trivial Hurewicz cofibration, though this takes a bit more work to see. The hard part is proven in prop 7 of these notes, as well as many textbooks.
Now let us first consider the special case of your conjecture where $f: B' \to B$ is a trivial Hurewicz cofibration.
Lemma: If $f: B' \to B$ is a trivial Hurewicz cofibration and $E \to B$ is a Hurewicz fibration, then we get an induced bijection:
$$ f^*: \pi_0 \Gamma(B, E) \to \pi_0 \Gamma(B', f^*E) $$
Proof: First let's show surjectivity. A section of $f^*E$ is the same as a map $s: B' \to E$ such that $ps = f$. This is a triangle which we can enlarge into a square where the left edge is $f$, the right edge is $p: E \to B$ and the bottom is the identity on $B$. This square is a "lifting problem". Now we use the property that trivial Hurewicz cofibrations have the left lifting property with respect to Hurewicz fibrations to solve the lifting problem. This solution is a map $B \to E$ which is exactly a section of $E$ which restricts on $B'$ to the original section.
Injectivity is proven by the same argument but using the fact that
$$ B \times \{0,1\} \cup^{B' \times \{0,1\}} B' \times I \to B \times I$$
is also a trivial Hurewicz cofibration. [We leave this part as an exercise]. QED.
Now using this lemma we can prove the general conjecture as follows. We will construct a space Z with two trivial Hurewicz cofibrations
$$ i:B \to Z $$
$$ j:B' \to Z $$
and a map $k:Z \to B$ such that the composite
$$B \to Z \to B $$
is the identity map on $B$ and the composite
$$B' \to Z \to B $$
is our map $f$. Once we have a space with these maps, the previous lemma shows that the maps between homotopy classes of sections $i^*$ and $j^*$ are bijections. Since $k^* i^* = 1^*$ is also a bijection, this means that $k^*$ is a bijection too, and hence $f^* = k^* j^*$ is a bijection, which is what we wanted to show.
Such a space $Z$ can be constructed as
$$ B' \times [0,1] \cup^{B' \times 1} B \times [1,2] $$
that is we glue $B' \times I$ to $B \times I$ at one end via the map $f$. The maps $i,j$ are given by including $B'$ and $B$ at 0 and 2, respectively (the ends of the "cylinder"). The map $k$ is given by projecting $B$ (by using $f$ for points on the first half of the cylinder).
Best Answer
The problem comes from the fact that $S^3\subset S^6$ admits more $PL$-framings than smooth ones: $\pi_3 SO_3=\mathbb{Z}$, while $\pi_3 PL_{6,3}=\mathbb{Z}^2$. (The group $PL_{6,3}$ is the group of PL self homeomorphisms of $\mathbb{R}^6$ preserving $\mathbb{R}^3\subset\mathbb{R}^6$ pointwise.) Thus, when we PL isotop the Haefliger trefoil identically to the trivial knot, and we lift this isotopy to its tubular neighborhood $T=S^3\times D^3$, the obtained $PL$-framing of the trivial knot is not smoothable i.e., it does not come from an $SO_3$-framing. In both smooth and PL categories the space of embeddings $T\hookrightarrow S^6$ that send $S^3\times *$ identically to itself is homotopy equivalent to the space of such immersions, while the latter space by Smale-Hirsch-Haefliger-Poenaru is equivalent to $Map(S^3,SO_3)$ and $Map(S^3,PL_{6,3})$, respectively.