Here is an example that shows that you can expect things to get as bad as it goes (I learned about this algebra from the wonderful article The Non-Commutative Gröbner Freaks by Green, Mora, and Ufnarovski who attribute it to Backelin).

The algebra $A$ has three generators $x,y,z$, and impose the relations $x^2=0, xy+zx=0$. The algebra $B$ is the subalgebra of $A$ generated by $x,y$. To determine the defining relations of $B$, we may compute the Gröbner basis for the weighted graded lexicographic ordering where the weight of $z$ is $1$ and the weights of $x,y$ are $0$. Then the weight $0$ part of the Gröbner basis is clearly going to be the Gröbner basis of the ideal of relations of $B$. Now, an easy calculation shows that the reduced Gröbner basis for this ordering consists of the elements $$ xy+zx, \ xy^kx\ \ (k\ge 0), $$ and so the subalgebra $B$ is not only not quadratic, but its minimal set of relations is infinite.

EDIT: following a comment of BugsBunny, here is an example of a Koszul algebra $A$ and its non-quadratic subalgebra $B$: let $A$ have three generators $x,y,z$, and relations $xy-yx=z^2, yz-zy=x^2, zx-xz=y^2$. This algebra (known as one of the Sklyanin algebras) is Koszul (see Artin, Tate, van den Bergh, "Some Algebras Associated to Automorphisms of Elliptic Curves"), and its component of dimension $n$ has the same dimension as the component of dimension $n$ of the algebra of polynomials in three variables. However, if we consider the subalgebra $B$ generated by $x,y$, we see that it has no quadratic relations, so if it were quadratic, it would be free and hence would have exponential growth of components, which is a contradiction.

Here is an example that shows that you can expect things to get as bad as it goes (I learned about this algebra from the wonderful article The Non-Commutative Gröbner Freaks by Green, Mora, and Ufnarovski who attribute it to Backelin).

The algebra $A$ has three generators $x,y,z$, and impose the relations $x^2=0, xy+zx=0$. The algebra $B$ is the subalgebra of $A$ generated by $x,y$. To determine the defining relations of $B$, we may compute the Gröbner basis for the weighted graded lexicographic ordering where the weight of $z$ is $1$ and the weights of $x,y$ are $0$. Then the weight $0$ part of the Gröbner basis is clearly going to be the Gröbner basis of the ideal of relations of $B$. Now, an easy calculation shows that the reduced Gröbner basis for this ordering consists of the elements $$ xy+zx, \ xy^kx\ \ (k\ge 0), $$ and so the subalgebra $B$ is not only not quadratic, but its minimal set of relations is infinite.

EDIT: following a comment of BugsBunny, here is an example of a Koszul algebra $A$ and its non-quadratic subalgebra $B$: let $A$ have three generators $x,y,z$, and relations $xy-yx=z^2, yz-zy=x^2, zx-xz=y^2$. This algebra (known as one of the Sklyanin algebras) is Koszul (see Artin, Tate, van den Bergh, "Some Algebras Associated to Automorphisms of Elliptic Curves"), and its component of dimension $n$ has the same dimension as the component of dimension $n$ of the algebra of polynomials in three variables. However, if we consider the subalgebra $B$ generated by $x,y$, we see that it has no quadratic relations, so if it were quadratic, it would be free and hence would have exponential growth of components, which is a contradiction.

EDIT2: here is a commutative version of the previous result for the reader curious about subalgebras of commutative Koszul algebras. Let $A$ be the commutative algebra with three generators $x,y,z$, and relations $xy=z^2, yz=x^2, zx=y^2$. Then the reduced Gröbner basis of relations of $A$ for the ordering $x>y>z$ consists of the four elements $x^2 - yz$, $xy - z^2$, $xz - y^2$, $y^3 - z^3$, which instantly implies that the subalgebra $B$ generated by $y,z$ has just one relation $y^3=z^3$. In fact, the algebra $A$ can be shown to be Koszul (for example, its [noncommutative] Koszul dual algebra has a quadratic Gröbner basis).

Here is an example that shows that you can expect things to get as bad as it goes (I learned about this algebra from the wonderful article The Non-Commutative Gröbner Freaks by Green, Mora, and Ufnarovski who attribute it to Backelin).

The algebra $A$ has three generators $x,y,z$, and impose the relations $x^2=0, xy+zx=0$. The algebra $B$ is the subalgebra of $A$ generated by $x,y$. To determine the defining relations of $B$, we may compute the Gröbner basis for the weighted graded lexicographic ordering where the weight of $z$ is $1$ and the weights of $x,y$ are $0$. Then the weight $0$ part of the Gröbner basis is clearly going to be the Gröbner basis of the ideal of relations of $B$. Now, an easy calculation shows that the reduced Gröbner basis for this ordering consists of the elements $$ xy+zx, \ xy^kx\ \ (k\ge 0), $$ and so the subalgebra $B$ is not only not quadratic, but its minimal set of relations is infinite.

Here is an example that shows that you can expect things to get as bad as it goes (I learned about this algebra from the wonderful article The Non-Commutative Gröbner Freaks by Green, Mora, and Ufnarovski who attribute it to Backelin).

The algebra $A$ has three generators $x,y,z$, and impose the relations $x^2=0, xy+zx=0$. The algebra $B$ is the subalgebra of $A$ generated by $x,y$. To determine the defining relations of $B$, we may compute the Gröbner basis for the weighted graded lexicographic ordering where the weight of $z$ is $1$ and the weights of $x,y$ are $0$. Then the weight $0$ part of the Gröbner basis is clearly going to be the Gröbner basis of the ideal of relations of $B$. Now, an easy calculation shows that the reduced Gröbner basis for this ordering consists of the elements $$ xy+zx, \ xy^kx\ \ (k\ge 0), $$ and so the subalgebra $B$ is not only not quadratic, but its minimal set of relations is infinite.

EDIT: following a comment of BugsBunny, here is an example of a Koszul algebra $A$ and its non-quadratic subalgebra $B$: let $A$ have three generators $x,y,z$, and relations $xy-yx=z^2, yz-zy=x^2, zx-xz=y^2$. This algebra (known as one of the Sklyanin algebras) is Koszul (see Artin, Tate, van den Bergh, "Some Algebras Associated to Automorphisms of Elliptic Curves"), and its component of dimension $n$ has the same dimension as the component of dimension $n$ of the algebra of polynomials in three variables. However, if we consider the subalgebra $B$ generated by $x,y$, we see that it has no quadratic relations, so if it were quadratic, it would be free and hence would have exponential growth of components, which is a contradiction.