Can intuitionistic proofs be laid out in a format that parallels reductio ad absurdum? Suppose I want to prove $A \Rightarrow B$ using classical logic. I can structure my proof in the following way.

Suppose not (If $A$, then $B$.)

Then $A$, and not $B$.
So...
...
Therefore, contradiction.

Now the above approach isn't valid intuitionistically (because $\neg(A\Rightarrow B)$ does not necessarily imply $A \wedge \neg B$.) However, I was wondering if a similar approach can be used in intuitionistic proofs. I'm thinking that additionally to "not," we define a new operator "alternative-not," which allows us to argue as follows.

Suppose alternative-not (If $A$, then $B$.)

Then $A$, and alternative-not $B$.
So...
...
Therefore, we're done.

Does an operator "alternative-not" with the necessary properties exist?
 A: 1) The main thing to say the canonical way of proving a conditional is not, repeat not, the round-the-houses route via reductio which you suggest, but (of course) simple Conditional Proof:

$\quad\quad |\quad A \quad\quad\quad\mathrm{supposition}$
$\quad\quad |\quad \vdots $
$\quad\quad |\quad C$
$A \to C \quad\quad\quad\quad\mathrm{discharge\ supposition}$

And this rule applies equally in intuitionistic logic.
2) Reductio in the form "If the supposition $A$ implies the absurdity $\bot$, then infer $\neg A$" is, repeat is, intuitionistically valid. 
3) The move from not-(If A, then B) to A and not-B is not inituitionistically valid. One way to see why it isn't intuitionistically valid is to reflect how we warrant that move in a natural deduction setting? 

$\neg(A \to B)$
$\quad | \quad \neg(A \land \neg B)$
$\quad | \quad | \quad A$
$\quad | \quad | \quad | \quad \neg B$
$\quad | \quad | \quad | \quad (A \land\neg B)$
$\quad | \quad | \quad | \quad \bot$
$\quad | \quad | \quad \neg\neg B$
$\quad | \quad | \quad B\quad\quad\quad$(DN)
$\quad | \quad (A \to B)\quad\quad$(CP)
$\quad | \quad \bot$
$\neg\neg(A \land \neg B)$
$(A \land \neg B)\quad\quad\quad$(DN)

This classical proof involves applications of the classical DN rule, so can't carry over to intuitionistic logic.
4) "Weak not"??? Since not-(If A, then B) to A and not-B is not intuitionistically valid, it looks more like you to introduce a stronger (indeed, classical!) not. What would the inference rules for this alternative negation be? How would it relate to the $\bot$ that appears in the essential intuitionistic rules (i) from $\bot$ infer anything, and (ii) reductio?
A: An identity is true in all Heyting algebras if and only if it is true in the algebra of open subsets of $\mathbb{R}^2$. In this algebra, $U \wedge V= U \cap V$, $U \vee V= U \cup C$, $\neg U = \text{ext}(U)$ (the exterior of $U$), $U \to V = \text{int}(U^c \cap V)$, and $(U \leq V) = (U \subseteq V)$
As first-order intuitionistic logic is complete, the validity of any argument can be checked by looking at truth values.
For the purposes of finding counter-examples, you should remove interior points from your open sets. e.g. it's not too hard to show
$$ \neg \neg U = \text{int}(\overline{U}) $$
Reasoning topologically, it seems unlikely that anything of the sort you want is possible without expanding the set of truth values.
e.g. $U \to V$ remembers the "holes" in $V$ that are away from $U$, but $\neg V$ has completely forgotten that information, and it is implausible that any open-set-valued operator can usefully retain that information in a complemented form.

Incidentally, RAA is valid in intuitionistic logic. If you suppose $P$ and derive a contradiction, then that means you've proven $P \to \bot$, which is equivalent to $\neg P$ (in fact, it's often taken as the definition of $\neg P$ in Heyting algebra). What is different is that $\neg \neg P$ does not always imply $P$.
However, an interesting feature of intuitionistic logic is that if you restrict yourself solely to working with negated propositions, the rules of deduction are identical to classical logic. e.g. $\neg\neg\neg P \to \neg P$ is a tautology. More generally, if $\neg P \to \neg Q$ is a tautology of classical logic, then it is automatically a tautology in intuitionistic logic too.
