I checked some more reference and come up with the following idea,this is too long for a comment,so i present it as an "answer":

The obstruction to the lifting is sitting in $H^4(S^4,\pi_3(\widetilde{Diff}(S^3)/Diff(S^3)))$.

Considere the homotopy exact sequence of the fibration $$\widetilde{Diff}(S^3)/Diff(S^3)\to BDiff(S^3)\to B \widetilde{Diff}(S^3)$$

We have $$\cdots\to\pi_3(O(4))\xrightarrow{p} \pi_3(\widetilde{Diff}(S^3))\to \pi_3(\widetilde{Diff}(S^3)/Diff(S^3))\to \pi_2(O(4))\cdots$$

We know $\pi_2(O(4))=0$ and to compute $\pi_3(\widetilde{Diff}(S^3))$,we need another fibration

$$S(M)\to B\widetilde{Diff}(M)\to BG(M)$$

where $S(M)$ is the structure set of $M$ and $BG(M)$ is the classifying space of the monoid of self homotopy equivalence of $M$.since sphere is rigid,we know $S(M)$ is trivial,hence $$\pi_i(B\widetilde{Diff}(S^3))\cong \pi_i(BG(S^3))$$

If $p$ could be identified with the $J$-homomorphism (not very sure at this time),then it is a surjective homomorphism.This,together with the fact that $\pi_2(O(4))=0$ would imply $\pi_3(\widetilde{Diff}(S^3)/Diff(S^3))=0$,hence no obstruction to the lifting.i.e.every $S^3$ block bundle over $S^4$ admits a fiber bundle structure.

could this homomorphism $p$ really be identified with the $J$-homomorphism? why or why not?

I checked some more reference and come up with the following idea,this is too long for a comment,so i present it as an "answer":

(as is pointed out,there is a mistake in my original argument,where i used $S^{diff}(S^n)$ is trivial,which is not true.while the original problem still makes sense in the Top category,where we have $S^{Top}(S^n)$ is trivial and $TOP(S^3)\simeq O(4)$)

The obstruction to the lifting is sitting in $H^4(S^4,\pi_3(\widetilde{TOP}(S^3)/TOP(S^3)))$.

Considere the homotopy exact sequence of the fibration $$\widetilde{TOP}(S^3)/TOP(S^3)\to BTOP(S^3)\to B \widetilde{TOP}(S^3)$$

We have $$\cdots\to\pi_3(O(4))\xrightarrow{p} \pi_3(\widetilde{TOP}(S^3))\to \pi_3(\widetilde{TOP}(S^3)/TOP(S^3))\to \pi_2(O(4))\cdots$$

We know $\pi_2(O(4))=0$ and to compute $\pi_3(\widetilde{TOP}(S^3))$,we need another fibration

$$S(M)\to B\widetilde{TOP}(M)\to BG(M)$$

where $S(M)$ is the structure set of $M$ and $BG(M)$ is the classifying space of the monoid of self homotopy equivalence of $M$.since sphere is topologically rigid,we know $S(M)$ is trivial,hence $$\pi_i(B\widetilde{TOP}(S^3))\cong \pi_i(BG(S^3))$$

If $p$ could be identified with the $J$-homomorphism (not very sure at this time),then it is a surjective homomorphism.This,together with the fact that $\pi_2(O(4))=0$ would imply $\pi_3(\widetilde{TOP}(S^3)/TOP(S^3))=0$,hence no obstruction to the lifting.i.e.every $S^3$ block bundle over $S^4$ admits a topological fiber bundle structure.

could this homomorphism $p$ really be identified with the $J$-homomorphism? why or why not?

I checked some more reference and come up with the following idea,this is too long for a comment,so i present it as an "answer":

The obstruction to the lifting is sitting in $H^4(S^4,\pi_3(\widetilde{Diff}(S^3)/Diff(S^3)))$.

Considere the homotopy exact sequence of the fibration $$\widetilde{Diff}(S^3)/Diff(S^3)\to BDiff(S^3)\to B \widetilde{Diff}(S^3)$$

We have $$\cdots\to\pi_3(O(4))\xrightarrow{p} \pi_3(\widetilde{Diff}(S^3))\to \pi_3(\widetilde{Diff}(S^3)/Diff(S^3))\to \pi_2(O(4))\cdots$$

We know $\pi_2(O(4))=0$ and to compute $\pi_3(\widetilde{Diff}(S^3))$,we need another fibration

$$S(M)\to B\widetilde{Diff}(M)\to BG(M)$$

where $S(M)$ is the structure set of $M$ and $BG(M)$ is the classifying space of the monoid of self homotopy equivalence of M.since sphere is rigid,we know $S(M)$ is trivial,hence $$\pi_i(B\widetilde{Diff}(S^3))\cong \pi_i(BG(S^3))$$

If $p$ could be identified with the $J$-homomorphism (not very sure at this time),then it is a surjective homomorphism.This,together with the fact that $\pi_2(O(4))=0$ would imply $\pi_3(\widetilde{Diff}(S^3)/Diff(S^3))=0$,hence no obstruction to the lifting.i.e.every $S^3$ block bundle over $S^4$ admits a fiber bundle structure.

could this homomorphism $p$ really be identified with the $J$-homomorphism? why or why not?

I checked some more reference and come up with the following idea,this is too long for a comment,so i present it as an "answer":

The obstruction to the lifting is sitting in $H^4(S^4,\pi_3(\widetilde{Diff}(S^3)/Diff(S^3)))$.

Considere the homotopy exact sequence of the fibration $$\widetilde{Diff}(S^3)/Diff(S^3)\to BDiff(S^3)\to B \widetilde{Diff}(S^3)$$

We have $$\cdots\to\pi_3(O(4))\xrightarrow{p} \pi_3(\widetilde{Diff}(S^3))\to \pi_3(\widetilde{Diff}(S^3)/Diff(S^3))\to \pi_2(O(4))\cdots$$

We know $\pi_2(O(4))=0$ and to compute $\pi_3(\widetilde{Diff}(S^3))$,we need another fibration

$$S(M)\to B\widetilde{Diff}(M)\to BG(M)$$

where $S(M)$ is the structure set of $M$ and $BG(M)$ is the classifying space of the monoid of self homotopy equivalence of $M$.since sphere is rigid,we know $S(M)$ is trivial,hence $$\pi_i(B\widetilde{Diff}(S^3))\cong \pi_i(BG(S^3))$$

If $p$ could be identified with the $J$-homomorphism (not very sure at this time),then it is a surjective homomorphism.This,together with the fact that $\pi_2(O(4))=0$ would imply $\pi_3(\widetilde{Diff}(S^3)/Diff(S^3))=0$,hence no obstruction to the lifting.i.e.every $S^3$ block bundle over $S^4$ admits a fiber bundle structure.

could this homomorphism $p$ really be identified with the $J$-homomorphism? why or why not?