Here's how to create a second preimage with your modification.
We'll assume that the original message is at least $2r+c$ bits long, and consists of an initial message segment $M_0$ ($2r+c$ bits) and the rest of the message $M_1$.
After processing block $M_0$, the sponge state will be some setting, with the public $r$ bits at some value $R_0$ and the capacity $c$ bits at some value $C_0$.
Concatenate the values $R_0$ and $C_0$ to be an initial block, and then prepend that to $M_1$, giving $R_0 || C_0 || M_1$.
When hashing that, the sponge construction will take the $R_0$ and $C_0$ bits and set the initial Sponge state to be precisely what it was after the initial processing of $M_0$. Then, it'll process the rest of the message $M_1$; because the state is the same, it'll perform the same operations, and generate the same hash.
And, because the lengths of the two messages are different, the messages are different, and thus we have generated a second preimage.
The OP since asked about the security if the length of the input is fixed. Well, if the length of the output $n$ is at most $r$ (i.e. one squeeze cycle), then it is easy to generate preimages. The method is straight-forward; you start with the final state (with the known $n$ bits in the $r$ region), and arbitrarily select values for the other $r+c-n$ state bits. Then, you run the hash backwards (using the inverse permutation, and arbitrary settings for intermediate message blocks) until you get to the initial state, and then you select a message where the first $r+c$ bits are that initial state, and the rest of the message blocks are those arbitrary message blocks you selected.
If we have $n > r$, this still can be used to reduce the amount of work to find a preimage by a factor of $2^r$
The bottom line: the entire security of SHA-3 comes from the capacity bits; letting the adversary arbitrarily select them breaks things bad...