No, for the same reason as in the linked question: an SO sentence over a given class of structures can be translated to an MSO sentence over structures augmented with their Cartesian powers, which can in turn be translated to an MSO sentence over graphs due to the universality of graphs. Thus, if MSO on graphs collapses to $\Sigma^1_k$, then the polynomial hierarchy collapses to $\mathrm{PH}=\Sigma^\mathrm P_k$.
More precisely, let $\Phi$ be an SO sentence in a finite signature $L$, quantifying over $l$-ary relations. Assume for simplicity that $L$ is relational. Let $L'=L\cup\{P(x_1,\dots,x_l,y)\}$, and for any $L$-structure $\mathcal A=(A,\dots)$, let $f(\mathcal A)$ be the $L'$-structure with domain $A\mathbin{\dot\cup}A^l$ that carries the original $L$-structure on $A$, and where $P$ is interpreted as the graph of the partial function that maps an $l$-tuple of elements of $A$ to the corresponding element of $A^l$. Notice that $f$ is polynomial-time computable. Then there is an MSO sentence $\Phi'$ such that
$$\mathcal A\models\Phi\iff f(\mathcal A)\models\Phi'$$
for every $L$-structure $\mathcal A$.
Moreover, there is a polynomial-time function $g$ mapping $L'$-structures to graphs, and a $1$-dimensional FO interpretation $I$ such that for any $L'$-structure $\mathcal A'$, $I$ interprets $\mathcal A'$ in $g(\mathcal A')$. (There are many ways how to do that; see e.g. Hedrlín & Pultr and Miller for variants of this construction in various contexts.) Then $\Phi'^I$ is an MSO sentence on graphs (more precisely, $\mathrm{MSO}_1$, i.e., quantifying over sets of vertices) such that
$$\mathcal A\models\Phi\iff g(f(\mathcal A))\models\Phi'^I.$$
If $\Phi'^I$ is equivalent to an $\Sigma^1_k$ sentence (not necessarily MSO), then the right-hand side can be evaluated in $\Sigma^\mathrm P_k$.
Since every PH language can be expressed by an SO sentence over finite structures in a suitable signature, we obtain $\mathrm{PH}=\Sigma^\mathrm P_k$ if such a collapse holds for all MSO sentences over graphs.
