The question would be even more interesting if the factor $1-x^2-y^2-w^2-z^2$ was replaced by $4-x^2-y^2-w^2-z^2$. The reason is that a necessary condition for the existence of such $p,q$ is that whenever $xywz>1$, the last factor is also negative. And it turns out that $$(xywz>1)\Longrightarrow(x^2+y^2+w^2+z^2<4),$$ by AM-GM inequality.

Notice that if the answer is positive with this constant $4$, then it is so with your constant $1$.

Now the answer: $p$ and $q$ exist for the sharp constant $4$. You may take $$q=\frac1{16}(4+x^2+y^2+w^2+z^2),\qquad p=\frac1{16}(x^2+y^2+w^2+z^2)^2-xywz.$$ To see $p$ is a sum of squares, just notice $$48p=(x^2+y^2-w^2-z^2)^2+\cdots+4(x-y)^2(w+z)^2+\cdots,$$ where the first sum consists of $3$ terms, and the second one consists in $6$ terms.

The question would be even more interesting if the factor $1-x^2-y^2-w^2-z^2$ was replaced by $4-x^2-y^2-w^2-z^2$. The reason is that a necessary condition for the existence of such $p,q$ is that whenever $xywz>1$, the last factor is also negative. And it turns out that $$(xywz>1)\Longrightarrow(x^2+y^2+w^2+z^2<4),$$ by AM-GM inequality.

Notice that if the answer is positive with this constant $4$, then it is so with your constant $1$.

Now the answer: $p$ and $q$ exist for the sharp constant $4$. You may take $q=\frac1{16}(4+x^2+y^2+w^2+z^2)$ and $$p=\frac1{16}(x^2+y^2+w^2+z^2)^2-xywz.$$ To see $p$ is a sum of squares, just notice $$48p=(x^2+y^2-w^2-z^2)^2+\cdots+4(x-y)^2(w+z)^2+\cdots,$$ where the first sum consists of $3$ terms, and the second one consists in $6$ terms.

The question would be even more interesting if the factor $1-x^2-y^2-w^2-z^2$ was replaced by $4-x^2-y^2-w^2-z^2$. The reason is that a necessary condition for the existence of such $p,q$ is that whenever $xywz>1$, the last factor is also negative. And it turns out that $$(xywz>1)\Longrightarrow(x^2+y^2+w^2+z^2<4),$$ by AM-GM inequality.

Notice that if the answer is positive with this constant $4$, then it is so with your constant $1$.

Now the answer: $p$ and $q$ exist for the sharp constant $4$. You may take $$q=\frac1{16}(4+x^2+y^2+w^2+z^2),\qquad p=\frac1{16}(x^2+y^2+w^2+z^2)^2-xywz.$$ To see $p$ is a sum of squares, just notice $$48p=(x^2+y^2-w^2-z^2)^2+\cdots+4(x-y)^2(w+z)^2+\cdots,$$ where each sum consists of $3$ terms.

The question would be even more interesting if the factor $1-x^2-y^2-w^2-z^2$ was replaced by $4-x^2-y^2-w^2-z^2$. The reason is that a necessary condition for the existence of such $p,q$ is that whenever $xywz>1$, the last factor is also negative. And it turns out that $$(xywz>1)\Longrightarrow(x^2+y^2+w^2+z^2<4),$$ by AM-GM inequality.

Notice that if the answer is positive with this constant $4$, then it is so with your constant $1$.

Now the answer: $p$ and $q$ exist for the sharp constant $4$. You may take $$q=\frac1{16}(4+x^2+y^2+w^2+z^2),\qquad p=\frac1{16}(x^2+y^2+w^2+z^2)^2-xywz.$$ To see $p$ is a sum of squares, just notice $$48p=(x^2+y^2-w^2-z^2)^2+\cdots+4(x-y)^2(w+z)^2+\cdots,$$ where the first sum consists of $3$ terms, and the second one consists in $6$ terms.