will $H(S\mathbin\|\mathtt{00})\mathbin\|H(S\mathbin\|\mathtt{01})\mathbin\|H(S\mathbin\|\mathtt{02})\mathbin\|H(S\mathbin\|\mathtt{03})\mathbin\|\ldots$ have the same cryptographic strength of the seed used even if the seed size is larger than the hash output or its internal state?

If we model $H$ as an ideal hash, that is as a random function with fixed size output: essentially† yes. For that model of the hash, each segment $H(S\mathbin\|\mathtt{uv})$ of the overall result for a particular input $F$ is independent of the others segments for that particular input $F$, and of every other segments for other inputs $F$.

For $H$ a practical hash: not quite, at least because the hash structure limits the number of possible outputs for any fixed size of $S$. For SHA-256 or any Merkle–Damgård hash with block size much larger than the state/output size, the worst case is when $S$ is exactly a multiple of the block size of the hash (512-bit for SHA-256): the Merkle–Damgård structure "limits" the number of states of the hash after processing $S$ to $2^{256}$, the multiple instances of $H$ are in the same 256-bit state at this point, thus there are at most $2^{256}$ output states for the overall output.

On the other hand, that's purely theoretical: 256-bit entropy is plenty enough, SHA-256 generally aims at "only" 128-bit security, and if we want more there's SHA-512.

Note that if we use $H(\mathtt{00}\mathbin\|S)\mathbin\|H(\mathtt{01}\mathbin\|S)\mathbin\|H(\mathtt{02}\mathbin\|S)\mathbin\|H(\mathtt{03}\mathbin\|S)\mathbin\|\ldots$ we somewhat improve things, because the multiple hashes no longer are in the same state, thus there can be much more entropy in the overall output.

† Even a perfect hash is bound to reduce entropy, but that's only slightly, see this; and this effect is very small in the context of the question.