Filtered objects and associated graded objects Given a filtered object one can consider it's associated graded object. However, non-isomorphic objects can have the isomorphic associated graded objects. 
The easiest example I could think of is the following:
Consider the ring $\mathbb{Z}$ and the filtration
$$\mathbb{Z}\supset 2\mathbb{Z}\supset 4\mathbb{Z}\supset 8\mathbb{Z}\supset\dots\supset 2^n\mathbb{Z}\supset\dots $$The associated graded algebra is $$\text{gr}(\mathbb{Z})=\bigoplus_{n\geq 0} \frac{2^n\mathbb{Z}}{2^{n+1}\mathbb{Z}},$$
the product is defined as follows: If $a\in 2^n\mathbb{Z}$ and $b\in 2^m\mathbb{Z}$, then $(a+2^{+1}\mathbb{Z})(b+2^{m+1}\mathbb{Z}):=ab+2^{n+m+1}\mathbb{Z}$.
Now consider the ring $\mathbb{Z}\oplus \mathbb{Z}_2$ and consider the same filtration as above summed with $\mathbb{Z}_2$ on the right. Clearly the associated graded rings are isomorphic but $\mathbb{Z}$ and $\mathbb{Z}\oplus \mathbb{Z}_2$ are not.
This is kind of a stupid example as I just got it by taking direct sums. I would like to see non-trivial examples (not necessarily for rings but preferably for Hopf-algebras). Any suggestions are wellcome.
 A: For any two $R$ modules, $A$ and $B$, there are some extensions $M$ that fit into a long exact sequence of the form $0 \to A \to M \to B \to 0$. Up to equivalence they are classified by the $R$ module $Ext_R^1(A,B)$, which can be non-trivial. There is always the trivial extension $M = A \oplus B$.
Now, take the filtration $0 \subset A \subset M$. The associated graded is the trivial extension.
For a very concrete example, you can take $A = \mathbb{Z}/2Z$ $B = Z/2Z$ as $R = \mathbb{Z}$ modules. Then you can find two extensions: $Z /4Z$ and $Z / 2Z \oplus Z / 2Z$, and you can compute that $Ext_R^1(A,B) = Z / 2Z$. 
There are many examples you can construct along these lines.

I also want to point out that knowing that associated graded can be a significant source of information. Indeed, it is all the spectral sequences compute.
Speculative and overly complicated comments:
I would guess that spectral sequences can give (unnecessarily complicated) examples of rings where the associated graded are equal, but the rings are not. For example, you can try to look at cohomology rings of two fiber bundles with the same fiber and base, but with total spaces with non-isomorphic cohomology rings. I don't think it will be formal (or even true?) that the spectral sequences are isomorphic, but you would have to chase through the construction of the differentials. Or you could try to compare the cohomology ring of one of these spaces to the spectral sequence computation.
If you want Hopf algebras, you can possibly look at the cohomology ring of some fiber bundles of Lie groups, and compare the (Hopf?) algebra on the $E^{\infty}$ page to the cohomology ring of the total Lie group, which is a Hopf algebra in general because of the map on cohomology induced by multiplication $m: G \times G \to G$. (I'm not sure if the pages of the spectral sequence are differential graded Hopf algebras, or whatever, but I guess that they are. If it is true then the proof is formal....)
However, there are certainly simpler examples coming from purely algebraic extensions, without worrying about topology and spectral sequences.
A: Another example is the universal enveloping algebra $U_\Bbbk(L)$ of a $\Bbbk$-Lie algebra $L$ with the canonical filtration induced by the one on $T_\Bbbk(L)$ ($\Bbbk$ a field of characteristic $0$). By PBW theorem we know that $\mathrm{gr}(U_\Bbbk(L))\cong S_\Bbbk(L)$, the symmetric algebra over $L$ (which is graded), but $U_\Bbbk(L)\ncong S_\Bbbk(L)$ as algebras (just as coalgebras).
