A "non-abelian excision" statement for mapping out of a space Let $U \subset A \subset X$ be spaces (in the sense of homotopy theory). 
For every pointed space $Y$ restriction maps induce the following canonical map between mapping spaces:
$$fiber(Map(X,Y) \to Map(A, Y)) \to fiber(Map(X-U,Y) \to Map(A-U,Y))$$
Is this map a homotopy equivalence? If not, can we give necessary and sufficient conditions (on $X, A, U$ and $Y$) for when it is?
 A: It depends on what exactly you mean by "subspace" and "fiber". Let me put some theorems down for you:
Theorem: Let $U,V\subseteq X$ open subspaces. Then the following is a homotopy pushout square:
$$\require{AMScd}
\begin{CD}
U\cap V @>>> U\\
@VVV @VVV \\
V @>>> U\cup V
\end{CD}\,.$$
Proof: This is a particular case of proposition A.3.2 in Higher Algebra, applied to the cover $\{U,V\}$ of $U\cup V$.$\,.\square$
Lemma: The derived functor $\mathrm{Map}(-,Y)$ turns homotopy pushout squares into homotopy pullback squares.
Proof: Classical (for example Proposition 5.5.2.2 in Higher Topos Theory, but this must be already in Bousfield-Kan).$\,\square$
Lemma: Let 
$$\require{AMScd}
\begin{CD}
P @>{f}>> X\\
@VVV @VVV \\
Y @>{g}>> Z
\end{CD}\,.$$
be a homotopy pullback square where $Z$ is pointed. Then the map
$$\mathrm{hofib}(f)\to \mathrm{hofib}(g)$$
is an equivalence.
Proof: Complete the square to a cube where the back face is
$$\require{AMScd}
\begin{CD}
\mathrm{hofib}(f) @>>> *\\
@VVV @VVV \\
\mathrm{hofib}(g) @>>> *
\end{CD}\,.$$
Then this is a homotopy pullback square, since the top, bottom and front faces are homotopy pullback squares. Then the thesis follows immediately. $\,\square$
Putting the two previous lemmas together you see that what you want is true (at least regarding the homotopy fibers) if the square
$$\require{AMScd}
\begin{CD}
A\smallsetminus U @>>> A\\
@VVV @VVV \\
X\smallsetminus U @>>> X
\end{CD}\,.$$
is a homotopy pushout square. This happens if $X$ is a simplicial complex and  $A$ and $X\smallsetminus U$ are subcomplexes, or (by the previous theorem) if $A$ is open and $U$ is closed.
