Are Blind Signatures Fundamentally Different From Signatures of Hashed Values?
The idea behind a blind signature is that party $\mathcal{A}$ has a message $m$ that they want party $\mathcal{B}$ to sign, but they do not want party $\mathcal{B}$ to learn the value of $m$. Using RSA, where $(e, N)$ is $\mathcal{B}$'s public key and $d$ is it's private key, this may look like:


$\mathcal{A}$ computes $x = m*r^e \mod N$, where $r$ is a randomly sampled value.
$\mathcal{B}$ signs $x$ i.e. $y = x^d \mod N$ and sends the signature to $\mathcal{A}$.
$\mathcal{A}$ computes $s \equiv y * r^{-1} \equiv (m * r^e)^d * r^{-1} \equiv m^d * r * r^{-1} \equiv m^d \mod N$, which is $\mathcal{B}$'s signature of $m$.


Note that $\mathcal{A}$ now has a signature from $\mathcal{B}$ on $m$, and $\mathcal{B}$ has no knowledge of $m$ since they cannot recover $m$ from $x$. [1] Anyone can verify this signature using $\mathcal{B}$'s public key.



Note that this does not work with hashing, because hashing is one-way. If we say $x = H(m)$ then in step 3. we have no way of recovering the signature for the original value $m$. 

[1]: as poncho points out, blind signatures have a stronger property than this, namely that if $\mathcal{B}$ signs messages for both $\mathcal{A}$ and $\mathcal{C}$, then given an unblinded message and signature pair $\mathcal{B}$ cannot determine who the signature was for