I interpret your question as basically asking "can we write a formula F logically equivalent to ¬(p→q) where F 1. is a conditional and 2. has a negation somewhere inside the conditional?" The answer is yes. I'll try to explain how I saw this first before providing an actual formula which does this.
First, since this question asks about the primary connective of a logical statement, let's rewrite all formulas in prefix notation. In other words, let's rewrite all formulas such that all logical operations precede their arguments (e. g. (p→q) becomes →pq). The first symbol for a formula in prefix notation indicates the type of statement. So, our problem becomes to find a formula with "→" as its first symbol with a negation symbol somewhere else in the formula.
Now implication and negation together form a set of adequate connectives for (two-valued) propositional logic. In other words, all formulas in (two-valued) propositional logic come as equivalent, in prefix notation, to a formula which has a conditional symbol or a negation symbol as its first symbol. Now The Sheffer Stroke "D", or Alternative Denial (NAND), by itself consists of an adequate connective; where D can get defined by the following table with 0 symbolizing falsity and 1 symbolizing truth:
D 0 1
0 1 1
1 1 0
The following logical equivalences (==) hold: 1. Dxx == ¬x, 2. Dxy == →x¬y. Now, since all formulas come as equivalent to a conditional or a negation, and every conditional or negation comes as equivalent to a Sheffer Stroke formula, and every Sheffer Stroke formula comes as equivalent to a conditional with a negation somewhere inside that conditional by 2., it follows that all formulas in classical propositional logic come as equivalent to such a conditional. One could make a similar argument using Peirce's arrow, or joint denial (NOR).
So, how might "¬(p→q)" or equivalently "¬→pq" look as a conditional? Well, substitute →pq for x in 1. above and we have ¬→pq comes as logically equivalent to D→pq→pq. Next, substituting →pq for both x and for y in 2. above we have D→pq→pq as logically equivalent to →→pq¬→pq. In infix notation that goes ((p→q)→(¬(p→q))).