Update: I believe this does not degrade security
Summary: For reasonable choice of basis for the Pedersen hash $\mathsf{PH}$, it is the case that
$$
\mathsf{xcoord}(\mathsf{PH}(\mathbf a)) = \mathsf{xcoord}(\mathsf{PH}(\mathbf a'))
\implies
\mathsf{PH}(\mathbf a) = \mathsf{PH}(\mathbf a').
$$
In other words, dropping the y-coordinate doesn't give you any advantage in finding collisions.
First, some math. What happens algebraically when you flip the y-coordinate of a Twisted Edwards curve? Recall the addition law for a curve of the form $ax^2 + y^2 = 1 + dx^2y^2$:
$$
(x_1, y_1) + (x_2, y_2) = \left(
\frac{x_1y_2 + y_1x_2}{1 + dx_1x_2y_1y_2},
\frac{y_1y_2 - ax_1x_2}{1 - dx_1x_2y_1y_2}
\right).
$$
Let $g(H)$ map a curve point $H = (x,y)$ to its x-reflection $(x, -y)$. Observe that, for any $H$,
$$
H + g(H)
= (x, y) + (x, -y) = \left(
\frac{xy - yx}{1 + dx^2y^2},
\frac{y^2 - ax^2}{1 - dx^2y^2}
\right)
= (0,-1)
=: Q.
$$
Thus, $g(H) = Q - H$ for all $H$. Also note, $Q$ has order 2, i.e., $Q + Q = \mathcal{O}$.
This is helpful for our characterization of collisions. Since the curve intersects any vertical line in at most two points, we have that for any two curve points $H \neq H'$,
$$
\mathsf{xcoord}(H) = \mathsf{xcoord}(H')
\iff H = g(H')
\iff H = Q - H'.
$$
What happens if we try to get an "easy" hash collision, i.e., getting a collision in the x-coordinate but not the y-coordinate? Let $\mathsf{PH}$ denote Pedersen hash function, mapping $k$ scalars $a_1, \ldots, a_k \in \mathbb F$ to $a_1 G_1 + \ldots + a_kG_k$, where $\{G_i\}$ denotes the Pedersen basis. An "easy" collision occurs with inputs $\mathbf{a}, \mathbf a'$ when $\mathsf{PH}(\mathbf a) \neq \mathsf{PH}(\mathbf a')$, and
\begin{align*}
&\mathsf{xcoord}(\mathsf{PH}(\mathbf a)) = \mathsf{xcoord}(\mathsf{PH}(\mathbf a'))
\\&\iff \mathsf{PH}(\mathbf a) = Q - \mathsf{PH}(\mathbf a')
\\&\iff {\textstyle\sum} a_iG_i = Q - {\textstyle\sum} a'_iG_i
\\&\iff {\textstyle\sum} (a_i + a'_i)G_i = Q.
\end{align*}
For certain choices of Pedersen basis $\{G_i\}$, this is a contradiction! If you select each $G_i$ from the large prime-order subgroup (a reasonable thing to do, since torsion is a headache in many protocols anyway), then you have that the LHS of the last equality is in the subgroup and the RHS is not. So if $(x,y)$ is an output of a $\mathsf{PH}$ with this kind of basis, then $(x, -y)$ cannot be an output of $\mathsf{PH}$.
Thus, the only way you can get an x-coordinate collision $\mathsf{xcoord}(\mathsf{PH}(\mathbf a)) = \mathsf{xcoord}(\mathsf{PH}(\mathbf a'))$ is to get a real collision $\mathsf{PH}(\mathbf a) = \mathsf{PH}(\mathbf a')$
Old Answer (ignore)
Outputting only the x-coordinate makes it trivial to find a collision.
Let $G_1, G_2, \ldots, G_k \in \mathbb G$ represent the Pedersen basis. The input to the Pedersen hash function $\mathsf H$ is $k$ scalars $a_1, \ldots, a_k \in \mathbb F$ and is mapped to the x-coordinate of $a_1G_1 + \ldots a_kG_k$.
Recall for any elliptic curve point $H = (x,y)$, it is the case that $-H = (x, -y)$. In other words, you get the negative by reflecting across the x-axis. The upshot is that for all $H$, $\mathsf{xcoord}(H) = \mathsf{xcoord}(-H)$.
Applying this to Pedersen, we get a collision:
$$
\mathsf H(a_1, \ldots, a_k)
=
\mathsf{xcoord}(a_1G_1 + \ldots a_kG_k)
=
\mathsf{xcoord}(-(a_1G_1 + \ldots a_kG_k))
=
\mathsf H(-a_1, \ldots, -a_k)
$$
