No. Suppose that $X$ is an $n$-sphere, $A$ is a closed hemisphere, and $U$ is a point in the interior of $A$. Then $SX$ is an $(n+1)$-sphere, $SA$ is a closed hemisphere, and $SU$ is a closed arc in $SA$ with endpoints in the boundary. $SX-SU$ is contractible and $SA-SU$ is homotopy-equivalent to an $(n-1)$-sphere, so $H(SX-SU,SA-SU)$ is in dimension $n$ whereas $H(SX,SA)$ is in dimension $n+1$.
EDIT: But maybe the following is what you wanted, or should have wanted. If $X=A\cup B$ and $C=A\cap B$ then we sometimes call the triad $(X;A,B)$ excisive if $H(B,C)\to H(X,A)$ is an isomorphism. As long as $S(A\cup B)=SA\cup SB$ and $S(A\cap B)=SA\cap SB$, the triad $SX;SA,SB)$ (SX;SA,SB)$will inherit the excision property from the original triad. To the question in the last paragraph: yes. 2 added 345 characters in body No. Suppose that$X$is an$n$-sphere,$A$is a closed hemisphere, and$U$is a point in the interior of$A$. Then$SX$is an$(n+1)$-sphere,$SA$is a closed hemisphere, and$SU$is a closed arc in$SA$with endpoints in the boundary.$SX-SU$is contractible and$SA-SU$is homotopy-equivalent to an$(n-1)$-sphere, so$H(SX-SU,SA-SU)$is in dimension$n$whereas$H(SX,SA)$is in dimension$n+1$. EDIT: But maybe the following is what you wanted, or should have wanted. If$X=A\cup B$and$C=A\cap B$then we sometimes call the triad$(X;A,B)$excisive if$H(B,C)\to H(X,A)$is an isomorphism. As long as$S(A\cup B)=SA\cup SB$and$S(A\cap B)=SA\cap SB$, the triad$SX;SA,SB)$will inherit the excision property from the original triad. To the question in the last paragraph: yes. 1 No. Suppose that$X$is an$n$-sphere,$A$is a closed hemisphere, and$U$is a point in the interior of$A$. Then$SX$is an$(n+1)$-sphere,$SA$is a closed hemisphere, and$SU$is a closed arc in$SA$with endpoints in the boundary.$SX-SU$is contractible and$SA-SU$is homotopy-equivalent to an$(n-1)$-sphere, so$H(SX-SU,SA-SU)$is in dimension$n$whereas$H(SX,SA)$is in dimension$n+1\$.