Since I am most familiar with the (co)chain complex version (using the surjection operad (cf. McClure-Smith)), I will give an example in this setting. The surjection operad $S$ is also filtered by the chain operads $S_k$ consisting of all surjections of complexity $\le k$ (as define in the paper above).

Playing a bit around with the definition of complexity shows that all operations of complexity $\le 1$ are just iterated applications of the $\cup$-product (which is also called $\langle 12\rangle$) and possibly permuting the arguments. So a cochain complex with the structure of an $S_{1}$-algebra is just a DGA.

Now the operation $\cup_1=\langle 121\rangle$ lives in $S_{2}$. It is a reason why the cohomology ring of our DGA is graded commutative, i.e. for two cocycles $x,y$ we have $$d(x\cup_1 y)=x \cup y \pm y\cup x.$$ So given a DGA whose cohomology ring is not graded commutative, cannot be quasiisomorphic to a DGA with the additional structure of an $S_2$-algebra, i.e. there is no map of DGAs which induces an isomorphism in cohomology.

Similarly, in the case of $\mathbb{F}_2$-coefficients, one can do the following to get obstructions to extend an $S_2$-algebra structure to an $S_3$-algebra. Suppose we have an $S_2$-algebra $C^*$. Especially we have $\cup$ and $\cup_1$-products so we can define for $[x]\in H^n(C^*)$ the elements $Sq^n([x])=[x\cup x]\in H^{2n}(C^*)$ and $Sq^{n-1}(x)=[x\cup_1 x]\in H^{2n-1}(C^*)$. Now if we have an Adem- or a Cartan relation that only involves these Operations, we can try to find a proof of those relations via the surjection operad, i.e. find a linear combination of operations whose boundary is the difference of the representatives of both sides (analogous to the case of graded commutativity above). Now those operations typically live in $S_k$ for some $k> 2$, and we end up with an obstruction to lifting the $S_2$-algebra structure to an $S_k$-algebra structure. The easiest example of this is probably that $Sq^1Sq^1=0:H^1\rightarrow H^3.$

So given a cochain complex over $\mathbb{F}_2$ with the structure of an $S_2$-algebra not satisfying that $Sq^1Sq^1=0:H^1\rightarrow H^3$, then the $S_2$-algebra structure cannot be lifted to an $S_3$-algebra structure. These are of course just a couple of examples of rather explicit obstructions. I have no idea whether this strategy is in any sense exhaustive, i.e. whether all obstructions can be constructed in a similar way.

Since I am most familiar with the (co)chain complex version (using the surjection operad (cf. McClure-Smith)), I will give an example in this setting. The surjection operad $S$ is also filtered by the chain operads $S_k$ consisting of all surjections of complexity $\le k$ (as define in the paper above).

Playing a bit around with the definition of complexity shows that all operations of complexity $\le 1$ are just iterated applications of the $\cup$-product (which is also called $\langle 12\rangle$) and possibly permuting the arguments. So a cochain complex with the structure of an $S_{1}$-algebra is just a DGA.

Now the operation $\cup_1=\langle 121\rangle$ lives in $S_{2}$. It is a reason why the cohomology ring of our DGA is graded commutative, i.e. for two cocycles $x,y$ we have $$d(x\cup_1 y)=x \cup y \pm y\cup x.$$ So given a DGA whose cohomology ring is not graded commutative, cannot be quasiisomorphic to a DGA with the additional structure of an $S_2$-algebra, i.e. there is no map of DGAs which induces an isomorphism in cohomology.

Similarly, in the case of $\mathbb{F}_2$-coefficients, one can do the following to get obstructions to extend an $S_2$-algebra structure to an $S_3$-algebra. Suppose we have an $S_2$-algebra $C^*$. Especially we have $\cup$ and $\cup_1$-products so we can define for $[x]\in H^n(C^*)$ the elements $Sq^n([x])=[x\cup x]\in H^{2n}(C^*)$ and $Sq^{n-1}(x)=[x\cup_1 x]\in H^{2n-1}(C^*)$. Now if we have an Adem- or a Cartan relation that only involves these Operations, we can try to find a proof of those relations via the surjection operad, i.e. find a linear combination of operations whose boundary is the difference of the representatives of both sides (analogous to the case of graded commutativity above). Now those operations typically live in $S_k$ for some $k> 2$, and we end up with an obstruction to lifting the $S_2$-algebra structure to an $S_k$-algebra structure. The easiest example of this is probably that $Sq^1Sq^1=0:H^1\rightarrow H^3.$

So given a cochain complex over $\mathbb{F}_2$ with the structure of an $S_2$-algebra not satisfying that $Sq^1Sq^1=0:H^1\rightarrow H^3$, then the $S_2$-algebra structure cannot be lifted to an $S_3$-algebra structure. These are of course just a couple of examples of rather explicit obstructions. I have no idea whether this strategy is in any sense exhaustive, i.e. whether all obstructions can be constructed in a similar way.

EDIT:

So what both examples have in common is that I am looking for a linear combination of operations with $m$ arguments of complexity $k$ such that its boundary has smaller complexity. So maybe we are really interested in $H^*((S_{k,r}/S_{m,r})\otimes_{RG}R)$, where $S_{k,r}$ is the cochain complex of all operations of complexity $\le k$ with $r$ arguments and $G$ is a subgroup of $\Sigma_r$ fixing a decomposition of $\{1,\ldots,r\}$ corresponding to plugging in the same argument multiple times. In the second example $G=\Sigma_4$. If we would look at a Cartan-relation, we would look at operations with four arguments and plug in something like $x\otimes x\otimes y\otimes y$ and get $G=\Sigma_2\times \Sigma_2\subset \Sigma_4$.

It also turns out that if we pick a boundary in the upper complex, then its boundary is strictly zero, i.e. the obstruction that we were hoping for holds on the nose in any $S_m$-algebra.