Yes, it is possible. (I assume you mean $f(E)$ with the reduced subscheme structure.)

Let $Y$ be the quintessential example of a non-Cohen-Macaulay surface: Take a smooth surface and glue two of its points "transversally". By this I mean that $Y$ along its only singular point would be analytically locally isomorphic to $Y'=Z(x,y)\cup Z(z,t)\subseteq \mathbb A^4_{x,y,z,t}$.

For the following computation this is what matters, so I will assume that $Y=Y'$. Blow up the singular point of $Y$: the two components (branches, really) separate and you get a $(-1)$-curve on each. Now glue the two surfaces together along their just created $(-1)$-curves to form a normal crossing singularity. This is $(-1)$-curve is your $E$ and the glued surfaces together is $X$ (if you started with an irreducible $Y$, then $X$ will also be irreducible). If you must have an $H$ as well, then blow up another point on $E$, but $H$ is really a red herring.

Anyway, the original blow up morphism on each component gives a morphism $f:X\to Y$ which contracts $E\simeq \mathbb P^1$ to a single point, that is, to the singular point of $Y=Y'$. However, if you blow up that point, then the two components (branches) get separated, so $f$ is not the blow-up of $f(E)$.

If you really want a $\mathbb P^n$-bundle with $\dim f(E)>0$, then just take a product of all of this with you favorite smooth variety. If you want $n>1$, then do this with higher dimensional smooth varieties meeting in a single point.

Yes, it is possible.

Well, it is kind of cheap, but if $f$ is not projective, then there is no chance of it being a blow-up, so you might as well say that $f$ is projective. Also, I assume you mean $f(E)$ with the reduced subscheme structure.

Let $Y$ be the quintessential example of a non-Cohen-Macaulay surface: Take a smooth surface and glue two of its points "transversally". By this I mean that $Y$ along its only singular point would be analytically locally isomorphic to $Y'=Z(x,y)\cup Z(z,t)\subseteq \mathbb A^4_{x,y,z,t}$.

For the following computation this is what matters, so I will assume that $Y=Y'$. Blow up the singular point of $Y$: the two components (branches, really) separate and you get a $(-1)$-curve on each. Now glue the two surfaces together along their just created $(-1)$-curves to form a normal crossing singularity. This glued together "double" $(-1)$-curve is your $E$ and the glued surfaces together is $X$ (if you started with an irreducible $Y$, then $X$ will also be irreducible). If you must have an $H$ as well, then blow up another point on $E$, but $H$ is really a red herring.

Anyway, the original blow up morphism on each component gives a morphism $f:X\to Y$ which contracts $E\simeq \mathbb P^1$ to a single point, that is, to the singular point of $Y=Y'$. However, if you blow up that point, then the two components (branches) get separated, so $f$ is not the blow-up of $f(E)$.

If you really want a $\mathbb P^n$-bundle with $\dim f(E)>0$, then just take a product of all of this with your favorite smooth variety. If you want $n>1$, then do this with higher dimensional smooth varieties meeting in a single point.

References for your questions in the comments below:

1. Gluing two surfaces along a curve should be pretty simple, but one could also just appeal to Karl Schwede's paper Gluing schemes and a scheme without closed points, especially Theorem 3.4 and Corollary 3.9.
2. See [Hartshorne, III.7.17].