So I am answering my own question. I have been thinking about this question and other related questions for months, and now I have some results.

For this question, if we have a positively curved 6-manifold $M$ with $SO(3)$-action with discrete fixed point set, then the orbit space $M/SO(3)$ is homeomorphic to a 3-ball. For a priori it is a simply connected compact 3-manifold with boundary, thus it is homeomorphic to 3-ball minus finitely many open disks. But it is also an Alexandrov space with positive curvature: now by the soul theorem, it has at most one boundary component; so it is either a 3-sphere or a 3-ball. Now by the slice representation of $SO(3)$ around the discrete fixed point, the orbit space has boundary and the boundary orbit types mainly consist of $SO(3)/SO(2)$.

Now I need a few extra assumptions. I need to assume that there are no exceptional orbits, or in other case, there are no finite non-trivial stabilizer groups. Under this assumption, I can show that the boundary 2-sphere of the orbit space has only 2 fixed points, and no other orbit types. The argument is as follows:

Take $S^1$-fixed point $M^{S^1}$ of this action, where $S^1$ is any maximal torus of $SO(3)$. The $S^1$-fixed point component $M^{S^1}_0$ above the boundary of the orbit space is a branched double cover of $S^2$, since in each $SO(3)/SO(2)$-orbit, the $SO(2)$-fixed point set is a set of 2 elements. But $M^{S^1}_0$ is also a 2-sphere since it is a totally geodesic submanifold of $M$, thus it also has positive curvature. By Riemann-Hurwicz formula, a branched double cover of $S^2$ over $S^2$ has 2 branched points. Thus we have 2 points on the boundary of the orbit space which are "more singular" than $SO(3)/SO(2)$. These could a priori be fixed points or $SO(3)/O(2)$. But from the representation of $O(2)$, the existence of $SO(3)/O(2)$ orbit will force the existence of finite non-trivial stabilizer group, violating my assumption. Thus the 2 branched points must be 2 fixed points.

Now our picture of the orbit space is clearer. It is a 3-ball, where principal isotropy group in the interior is trivial, and the boundary orbit types are $SO(3)/SO(2)$ and fixed points, and there are 2 fixed points. We have two cases. Case I: interior orbits are all $SO(3)$; case II: there is one singular orbit $SO(3)/SO(2)$ in the interior of the orbit space. There can't be more singular orbits in the interior, because by the q-extent argument, the total number of interior singular orbits and boundary fixed points is at most 3.

I can solve Case I. In this case, $M^6$ is a suspension of a 5-dim $SO(3)$-space $N^5$ whose orbit space corresponds to the equator disk of $M/SO(3)$. In other words, $N^5/SO(3)$ is a 2-disk with two orbit types $SO(3)$ and $SO(3)/SO(2)$, and all the singular orbits lie on the boundary. According to the "second classification theorem" in Bredon's book , Ch.5, Theorem 6.1 and Corollary 6.2, there are 2 such $SO(3)$-spaces, and they are $S^5$ and $S^2\times S^3$. $M^6$ is suspension of one of these two, but only suspension of $S^5$ is a manifold. Thus $M^6$ is homeomorphic to $S^6$. Case I is done.

For Case II and more general cases where one allows exceptional orbits, I don't know what to do so far. I can't get a homeomorphism classfication in these case, but still I can say something about the (co)homology groups, using tools like Mayer-Vietoris sequence.

So I am answering my own question. I have been thinking about this question and other related questions for months, and now I have some results.

For this question, if we have a positively curved 6-manifold $M$ with $SO(3)$-action with discrete fixed point set, then the orbit space $M/SO(3)$ is homeomorphic to a 3-ball. For a priori it is a simply connected compact 3-manifold with boundary, thus it is homeomorphic to 3-ball minus finitely many open disks. But it is also an Alexandrov space with positive curvature: now by the soul theorem, it has at most one boundary component; so it is either a 3-sphere or a 3-ball. Now by the slice representation of $SO(3)$ around the discrete fixed point, the orbit space has boundary and the boundary orbit types mainly consist of $SO(3)/SO(2)$.

Now I need a few extra assumptions. I need to assume that there are no exceptional orbits, or in other case, there are no finite non-trivial stabilizer groups. Under this assumption, I can show that the boundary 2-sphere of the orbit space has only 2 fixed points, and no other orbit types. The argument is as follows:

Take $S^1$-fixed point $M^{S^1}$ of this action, where $S^1$ is any maximal torus of $SO(3)$. The $S^1$-fixed point component $M^{S^1}_0$ above the boundary of the orbit space is a branched double cover of $S^2$, since in each $SO(3)/SO(2)$-orbit, the $SO(2)$-fixed point set is a set of 2 elements. But $M^{S^1}_0$ is also a 2-sphere since it is a totally geodesic submanifold of $M$, thus it also has positive curvature. By Riemann-Hurwicz formula, a branched double cover of $S^2$ over $S^2$ has 2 branched points. Thus we have 2 points on the boundary of the orbit space which are "more singular" than $SO(3)/SO(2)$. These could a priori be fixed points or $SO(3)/O(2)$. But from the representation of $O(2)$, the existence of $SO(3)/O(2)$ orbit will force the existence of finite non-trivial stabilizer group, violating my assumption. Thus the 2 branched points must be 2 fixed points.

Now our picture of the orbit space is clearer. It is a 3-ball, whose principal isotropy group in the interior is trivial, and the boundary orbit types are $SO(3)/SO(2)$ and fixed points. There are 2 fixed points. We have two cases: Case I: interior orbits are all $SO(3)$; Case II: there is one singular orbit $SO(3)/SO(2)$ in the interior of the orbit space. There can't be more singular orbits in the interior, because by the q-extent argument, the total number of interior singular orbits and boundary fixed points is at most 3.

I can solve Case I. In this case, $M^6$ is a suspension of a 5-dim $SO(3)$-space $N^5$ whose orbit space corresponds to the equator disk of $M/SO(3)$. In other words, $N^5/SO(3)$ is a 2-disk with two orbit types $SO(3)$ and $SO(3)/SO(2)$, and all the singular orbits lie on the boundary. According to the "second classification theorem" in Bredon's book , Ch.5, Theorem 6.1 and Corollary 6.2, there are 2 such $SO(3)$-spaces, and they are $S^5$ and $S^2\times S^3$. $M^6$ is suspension of one of these two, but only suspension of $S^5$ is a manifold. Thus $M^6$ is homeomorphic to $S^6$. Case I is done.

For Case II and more general cases where one allows exceptional orbits, I do not know what to do so far. I can not get a homeomorphism classfication in these cases, but still I can say something about the (co)homology groups, using tools like Mayer-Vietoris sequence.

So I'm answering my own question.. I've been thinking about this question and other related questions for months, and I had some results.

For this question, if we have a positively curved 6-manifold $M$ with $SO(3)$-action with discrete fixed point set, then the orbit space $M/SO(3)$ is homeomorphic to a 3-ball. A priori it is a simply connected compact 3-manifold with boundary, thus it is homeomorphic to 3-ball minus finitely many open disks. But it is also an Alexandrov space with positive curvature, by the soul theorem, it has at most one boundary component; so it is a 3-sphere or a 3-ball. By the slice representation of $SO(3)$ around the discrete fixed point, the orbit space has boundary and the boundary orbit types mainly consist of $SO(3)/SO(2)$.

Now I need a few further assumptions. I need to assume that there are no exceptional orbits, or in other case, there are no finite non-trivial stabilizer groups. Under this assumption, I can show that the boundary 2-sphere of the orbit space has only 2 fixed points, and no other orbit types. The argument is as follows:

Take $S^1$-fixed point $M^{S^1}$ of this action, where $S^1$ is any maximal torus of $SO(3)$. The $S^1$-fixed point component $M^{S^1}_0$ above the boundary of the orbit space is a branched double cover of $S^2$, since in each $SO(3)/SO(2)$-orbit, the $SO(2)$-fixed point set is a set of 2 elements. But $M^{S^1}_0$ is also a 2-sphere since it is a totally geodesic submanifold of $M$, thus it also has positive curvature. By Riemann-Hurwicz formula, a branched double cover of $S^2$ over $S^2$ has 2 branched points. Thus we have 2 points on the boundary of the orbit space which are "more singular" than $SO(3)/SO(2)$. These could a priori be fixed points or $SO(3)/O(2)$. But from the representation of $O(2)$, the existence of $SO(3)/O(2)$ orbit will force the existence of finite non-trivial stabilizer group, violating my assumption. Thus the 2 branched points must be 2 fixed points.

Now our picture of the orbit space is more clear. It is a 3-ball, where principal isotropy group in the interior is trivial, and the boundary orbit types are $SO(3)/SO(2)$ and fixed points, and there are 2 fixed points. We have two cases. Case I: interior orbits are all $SO(3)$; case II: there is one singular orbit $SO(3)/SO(2)$ in the interior of the orbit space. There can't be more singular orbits in the interior, because by the q-extent argument, the total number of interior singular orbits and boundary fixed points is at most 3.

I can solve Case I. In this case, $M^6$ is a suspension of a 5-dim $SO(3)$-space $N^5$ whose orbit space corresponds to the equator disk of $M/SO(3)$. In other words, $N^5/SO(3)$ is a 2-disk with two orbit types $SO(3)$ and $SO(3)/SO(2)$, and all the singular orbits lie on the boundary. According to the "second classification theorem" in Bredon's book , Ch.5, Theorem 6.1 and Corollary 6.2, there are 2 such $SO(3)$-spaces, and they are $S^5$ and $S^2\times S^3$. $M^6$ is suspension of one of these two, but only suspension of $S^5$ is a manifold. Thus $M^6$ is homeomorphic to $S^6$. Case I is done.

For Case II and more general cases where one allows exceptional orbits, I don't know what to do so far. I can't get a homeomorphism classfication in these case, but still I can say something about the (co)homology groups, using tools like Mayer-Vietoris sequence.

So I am answering my own question. I have been thinking about this question and other related questions for months, and now I have some results.

For this question, if we have a positively curved 6-manifold $M$ with $SO(3)$-action with discrete fixed point set, then the orbit space $M/SO(3)$ is homeomorphic to a 3-ball. For a priori it is a simply connected compact 3-manifold with boundary, thus it is homeomorphic to 3-ball minus finitely many open disks. But it is also an Alexandrov space with positive curvature: now by the soul theorem, it has at most one boundary component; so it is either a 3-sphere or a 3-ball. Now by the slice representation of $SO(3)$ around the discrete fixed point, the orbit space has boundary and the boundary orbit types mainly consist of $SO(3)/SO(2)$.

Now I need a few extra assumptions. I need to assume that there are no exceptional orbits, or in other case, there are no finite non-trivial stabilizer groups. Under this assumption, I can show that the boundary 2-sphere of the orbit space has only 2 fixed points, and no other orbit types. The argument is as follows:

Take $S^1$-fixed point $M^{S^1}$ of this action, where $S^1$ is any maximal torus of $SO(3)$. The $S^1$-fixed point component $M^{S^1}_0$ above the boundary of the orbit space is a branched double cover of $S^2$, since in each $SO(3)/SO(2)$-orbit, the $SO(2)$-fixed point set is a set of 2 elements. But $M^{S^1}_0$ is also a 2-sphere since it is a totally geodesic submanifold of $M$, thus it also has positive curvature. By Riemann-Hurwicz formula, a branched double cover of $S^2$ over $S^2$ has 2 branched points. Thus we have 2 points on the boundary of the orbit space which are "more singular" than $SO(3)/SO(2)$. These could a priori be fixed points or $SO(3)/O(2)$. But from the representation of $O(2)$, the existence of $SO(3)/O(2)$ orbit will force the existence of finite non-trivial stabilizer group, violating my assumption. Thus the 2 branched points must be 2 fixed points.

Now our picture of the orbit space is clearer. It is a 3-ball, where principal isotropy group in the interior is trivial, and the boundary orbit types are $SO(3)/SO(2)$ and fixed points, and there are 2 fixed points. We have two cases. Case I: interior orbits are all $SO(3)$; case II: there is one singular orbit $SO(3)/SO(2)$ in the interior of the orbit space. There can't be more singular orbits in the interior, because by the q-extent argument, the total number of interior singular orbits and boundary fixed points is at most 3.

I can solve Case I. In this case, $M^6$ is a suspension of a 5-dim $SO(3)$-space $N^5$ whose orbit space corresponds to the equator disk of $M/SO(3)$. In other words, $N^5/SO(3)$ is a 2-disk with two orbit types $SO(3)$ and $SO(3)/SO(2)$, and all the singular orbits lie on the boundary. According to the "second classification theorem" in Bredon's book , Ch.5, Theorem 6.1 and Corollary 6.2, there are 2 such $SO(3)$-spaces, and they are $S^5$ and $S^2\times S^3$. $M^6$ is suspension of one of these two, but only suspension of $S^5$ is a manifold. Thus $M^6$ is homeomorphic to $S^6$. Case I is done.

For Case II and more general cases where one allows exceptional orbits, I don't know what to do so far. I can't get a homeomorphism classfication in these case, but still I can say something about the (co)homology groups, using tools like Mayer-Vietoris sequence.